Physics Problems Q&A - Recent questions in Physics Problems
http://physics.qandaexchange.com/?qa=questions/physics-problems
Powered by Question2AnswerFind position and velocity at t=0?
http://physics.qandaexchange.com/?qa=3377/find-position-and-velocity-at-t-0
<p>Full question: </p>
<blockquote><p>An object moves with constant acceleration. At t= 2.50 s, the position of the object is x=2.00 m and its velocity is v= 4.50 m/s. At t= 7.00 s, v= -12.0 m/s. <br>
Find: <br>
(a) the position and the velocity at t= 0;<br>
(b) the average speed from 2.50s to 7.00 s, and <br>
(c) the average velocity from 2.50s to 7.00 s.</p>
</blockquote>
<p>I tried using the kinematic equations of motion for constant acceleration but my answers make no sense. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3377/find-position-and-velocity-at-t-0Tue, 05 Feb 2019 00:03:55 +0000What is the fraction of work per second by F is converted into heat.
http://physics.qandaexchange.com/?qa=3373/what-is-the-fraction-of-work-per-second-by-converted-into-heat
<blockquote><p>Two long parallel horizontal rails, a distance l apart and each <br>
has a resistance $\lambda$ per unit length are joined at one end by a <br>
resistance R. A perfectly conducting rod MN of mass m is free to<br>
slide along the rails without friction. There is a uniform magnetic <br>
field of induction B normal to the plane of paper and directed <br>
into the paper. A variable force F is applied to the rod MN such <br>
that, as the rod moves, a constant current i flows through R. <br>
What is the fraction of work per second by F is converted into heat.</p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HYQy1Oj.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3373/what-is-the-fraction-of-work-per-second-by-converted-into-heatWed, 30 Jan 2019 13:44:47 +0000Find minimum distance between dipoles
http://physics.qandaexchange.com/?qa=3372/find-minimum-distance-between-dipoles
<p><img src="https://cdn.pbrd.co/images/HYQsmTc.jpg" alt=""></p>
<p>My try :</p>
<p><img src="https://cdn.pbrd.co/images/HYQsHWk.jpg" alt=""><br>
<img src="https://cdn.pbrd.co/images/HYQsXTS.jpg" alt=""><br>
But answer does not match.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3372/find-minimum-distance-between-dipolesWed, 30 Jan 2019 13:32:57 +0000Force that one half of uniformly charged solid sphere exerts on other half - Griffith 2.43
http://physics.qandaexchange.com/?qa=3368/force-uniformly-charged-solid-sphere-exerts-other-griffith
<p><img src="https://cdn.pbrd.co/images/HYuG13c.png" alt=""></p>
<p>I got the result mentioned, by considering the electric field as $\dfrac{\rho\cdot \vec{r}}{3\epsilon_o}$, and integrating carefully.</p>
<p>As problem asks to calculate the force on one part due to other, but the electric field I'm using here is the result of complete solid non conducting sphere, so why are we considering $E$ of the northern hemisphere to calculate force <strong>it</strong> experience. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3368/force-uniformly-charged-solid-sphere-exerts-other-griffithMon, 28 Jan 2019 06:18:15 +0000Electric field intensity at the centre of a solid hemisphere with azimuthal volume density
http://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-azimuthal-volume-density
<p><a rel="nofollow" href="https://imgur.com/a/n0iohng">https://imgur.com/a/n0iohng</a></p>
<p>I am using Guru's Electromagneic Theory Fundamentals: <a rel="nofollow" href="https://www.amazon.ca/Electromagnetic-Field-Theory-Fundamentals-Singh/dp/0521116023">https://www.amazon.ca/Electromagnetic-Field-Theory-Fundamentals-Singh/dp/0521116023</a></p>
<p>For finding the electric field intensity, I am using the equation:</p>
<p><a rel="nofollow" href="https://imgur.com/a/RHS6SfY">https://imgur.com/a/RHS6SfY</a></p>
<p>I am unsure about what the value of r prime should be for a solid. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-azimuthal-volume-densitySun, 20 Jan 2019 03:09:55 +0000Rotating ball inside rotating cylinder.
http://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinder
<p>How do I proof: <img src="https://cdn.pbrd.co/images/HWY3y5g.png" alt=""> </p>
<p>Textbook solution :<br>
<img src="https://cdn.pbrd.co/images/HXsICHVJ.png" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinderFri, 18 Jan 2019 05:18:03 +0000Frequency modes of the rectangular shell
http://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shell
<p>This is task i received from my professor:</p>
<blockquote><p>The shapes of three natural modes having the frequencies $\omega_1, \omega_2, \omega_3$ of the rectangular shell are presented in the figure. The exciting pressure $p(t)$ applied uniformly all over the one side of the shell has the form $p(t) = Pe^{jωt}$. <br>
Make a sketch of the normal displacement of the gravity point of the shell against frequency, if the excitation frequency varies within bounds $0.5\omega_1< \omega <2\omega_1$ and static displacement of that point equals to $u_0$.</p>
</blockquote>
<p>Links to photo of frequency nodes (sorry for low quality)-><a rel="nofollow" href="https://ibb.co/b1nh3j9">1</a> .</p>
<p>Can somebody help me and tell me how i should get started?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shellMon, 14 Jan 2019 21:42:44 +0000tension in wires from which a block is suspended
http://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspended
<p><img src="https://cdn.pbrd.co/images/HVdTW5k.png" alt="1"></p>
<p>I am trying to figure out why the answer is T>W. Do you have any idea why? If so could you please explain. Thank you.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspendedSun, 06 Jan 2019 15:45:40 +0000How to solve this throwing ball off cliff question
http://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-question
<p><img src="https://cdn.pbrd.co/images/HVdO2s0.png" alt=""></p>
<p>For b) I was curious why i couldn't use $s = ut + \frac12at^2$ as my equation as i had all the necessary known values. Instead to get the correct answer i had to use $v=u + at$.</p>
<p>Why is this so? Thank you for your time.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-questionSun, 06 Jan 2019 15:45:23 +0000Time period of small oscillation- disk and particle system
http://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-system
<p><img src="https://cdn.pbrd.co/images/HUWMQA2.jpg" alt=""></p>
<p>Attempt: </p>
<p>$y_{COM} = \dfrac{R}{3} = \dfrac {25}3$</p>
<p>$I = \dfrac 12 (2)R^2 + 1R^2 = 2\times 25 = 50 $</p>
<p>Now, T is given by: $T =2\pi \sqrt{\dfrac{I}{mgl}}$ where l is the distance of the centre of mass from the centre of rotation</p>
<p>So $T =2\pi \sqrt{\dfrac{50}{3 \times 10 \times \dfrac{25}{3}}} = 2\pi\sqrt{\dfrac{1}{5}}$</p>
<p>No option :( </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-systemFri, 04 Jan 2019 22:33:35 +0000Wave optics- number of minimas.
http://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimas
<p><img src="https://cdn.pbrd.co/images/HUWujTQ.jpg" alt=""></p>
<p>Attempt: </p>
<p>Let m be the integer associated with $420 nm$<br>
and n be the integer associated with $540 nm$</p>
<p>$d \sin \theta = k \lambda $ $k\in Z$<br>
Clearly, </p>
<p>$m_{max} = 180$ </p>
<p>$n_{max} = 140 $</p>
<p>$(2m+1) \lambda_1 = (2n+1)\lambda_2$ (condition for dark fringes to overlap)</p>
<p>$\implies \dfrac{2m+1}{2n+1} = \dfrac 97$</p>
<p>$\implies m = \dfrac{1+7n +2n }{7}$</p>
<p>Hence, we obtain, for m to be an integer: $2n+1 = 7k$</p>
<p>$\implies n = \dfrac{7k-1}{2}$ where $k \in Z$</p>
<p>Now note that k must be odd since odd-1 = even </p>
<p>Thus, using $n \le 140$, $k_{max} = 39$</p>
<p>Now, we have to consider only odd values of k which are $1,3,...39$ = 20 numbers </p>
<p>Thus, we have 20 minimas on the upper side and 20 on the lower,<br>
Total minimas = $20+20 = 40$</p>
<p>But answer is $D$</p>
<p><strong>Question 1:</strong> What is wrong with my method? </p>
<p><strong>Question 2:</strong> Considering that this is a JEE Mains problem, what is the fastest 2 minute way to do it? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimasFri, 04 Jan 2019 22:01:11 +0000oblique collision of a rigid rod with a plane surface
http://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surface
<blockquote><p>A uniform rod AB of length L is released from rest with AB inclined at angle 60° with horizontal, with the closer end of the rod to the ground, at a height 49m from it. Assuming the collision of the rod with the ground to be perfectly elastic, calculate the height the centre of mass of the rod will rebound after impact (assume ground to be frictionless).</p>
</blockquote>
Physics Problemshttp://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surfaceTue, 01 Jan 2019 16:23:00 +0000Cylinder lying on conveyor belt
http://physics.qandaexchange.com/?qa=3323/cylinder-lying-on-conveyor-belt
<blockquote><p>You buy a bottle of water in the store and place it on the conveyor belt with the longitudinal axis perpendicular to the direction of movement of the belt. Initially, both the belt and the bottle are at rest. We can approach the bottle as one cylinder with radius $a$, mass $M$ and moment of inertia $I = M k^2$ ($k$ in units of length), in which the mass is not distributed uniformly. The speed of the belt at time $t$ is $V (t)$.<br>
a) Find an expression for the speed $v(t)$ of the centre of mass of the bottle.<br>
b) Explain why a bottle tends to start spinning on one moving band.</p>
</blockquote>
<p>What I have tried:</p>
<p>a) Having both longitudinal axis and belt's velocity vector perpendicular to each other <strong>on the same plane</strong> means that there will be rolling motion if we assume that static friction will be overcome. Let's disregard slippery effects as well.</p>
<p>Here I used an approach based on the fact that <strong>the bottle-belt-Earth system is nonisolated in terms of energy because the belt exerts an external force on the bottle, which means that it does work on the bottle.</strong> Therefore:</p>
<p>$$W = \Delta K_{tr} + \Delta K_{rot}$$</p>
<p>$$(F-f_f)d =v_{CM}^2 (\frac{I_{CM}}{a^2} + M)$$</p>
<p>$$v_{CM} = \sqrt{ \frac{2(F-f_f)d}{\frac{I_{CM}}{a^2} + M}} = \sqrt{ \frac{(F-f_f)d}{ M}}$$</p>
<p>The result I got does not seem to be incorrect; I checked dimensions and got $\frac{L}{T}$ .<strong>My question here is if I am right including the force F. I consider this force as a fictitious one, which is triggered by the non inertial frame in which the bottle is located: the conveyor belt.</strong> Note I am analysing the scenario from an inertial reference frame: The Earth (we regard it as inertial for known reasons). </p>
<p>b) If the static friction force was equal to the fictitious force F the cylinder <strong>would not spin</strong>. It would just move transitionally along the belt. Actually, if we were located on the belt, we would not see the cylinder move at all.</p>
<p><strong>In order to spin, the magnitude of F must exceed the magnitude of the maximum force of static friction. This friction force is recalled as force of kinetic friction.</strong></p>
<p><strong>EDIT</strong></p>
<p>From second (translation) Newton's law:</p>
<p>$$a_o= \frac{f}{M}$$</p>
<p>From second (rotation) Newton's law:</p>
<p>$$\tau = I \alpha = fa$$</p>
<p>$$k^2 \alpha= a_o a$$</p>
<p> $$\alpha = \frac{a_o a}{k^2}$$</p>
<p>Where $a_o$ is the acceleration of the cylinder measured from the ground.</p>
<p><strong>Assuming that the acceleration of the belt is constant</strong> ($a_b$):</p>
<p>$$ \frac{V}{t} = a_b$$</p>
<p>The acceleration of the cylinder with respect to the ground accounts for the acceleration of the belt and the angular acceleration of the cylinder (which has a negative sign because I considered the cylinder spinning clockwise i.e. the belt accelerating to the left and regarding that direction as the positive one). Therefore:</p>
<p>$$a_o = a_b - a\alpha$$</p>
<p>$$a_o = \frac{V}{t} -a\frac{a_o a}{k^2}$$</p>
<p>$$a_o = \frac{Vk^2}{(k^2 + a^2)t}$$</p>
<p>We know by kinematics that:</p>
<p>$$v_{cm} = \frac{Vk^2}{k^2 + a^2}$$</p>
<p>This makes sense to me. However I am confused because <strong>I was suggested that I should not assume that the belt moves with constant acceleration</strong> (please see kuruman's #2 comment <a rel="nofollow" href="https://www.physicsforums.com/threads/cylinder-lying-on-conveyor-belt.963471/">https://www.physicsforums.com/threads/cylinder-lying-on-conveyor-belt.963471/</a> ); Then how could I solve this problem?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3323/cylinder-lying-on-conveyor-beltSat, 22 Dec 2018 16:56:04 +0000floating cubical block
http://physics.qandaexchange.com/?qa=3319/floating-cubical-block
<blockquote><p>A cubical block of copper of side $10\ cm$ is floating containing mercury. Water is poured into the vessel so that the coper in the block just gets submerged. The height of the water column is?</p>
</blockquote>
<p>Solution goes like :</p>
<blockquote><p>Let height of water column be $h$, then $$\rho_w gh + \rho_{Hg} g(10-h)=\rho_{Cu} g 10$$</p>
</blockquote>
<p>I don't understand this, why are they writing pressure using Mercury?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3319/floating-cubical-blockThu, 20 Dec 2018 11:03:40 +0000Prove that the vector of acceleration always points towards the origin
http://physics.qandaexchange.com/?qa=3312/prove-that-vector-acceleration-always-points-towards-origin
<blockquote><p>A planet travels in an ellipse around the origin with the sun close to the origin. The planet's position in the ellipse is given by:</p>
<p>$x(t)=a\cos(kt)$</p>
<p>$y(t)=b\sin(kt)$</p>
<p>where $a>b$ and $k$ are constants and $0 ≤ t ≤ 2\pi$ ($t$ is a time variable).</p>
<p>The sun's position is given by $(x,y) = ( \sqrt{ a^2 − b^2}, 0)$.</p>
</blockquote>
<p>I've learnt that taking the derivative of the functions for position twice gives the functions for the acceleration. How would this look and how can I compare this answer with the functions for positions to show that the vector of acceleration is always pointed towards the origin?</p>
<p>In my attempt I've got that the acceleration is</p>
<p>$$a_x(t) = −ak^2cos(kt), a_y(t) = −bk^2sin(kt)$$</p>
<p>So the difference between position and acceleration is $−k^2$, provided that I'm correct so far. However I do not understand how this would show that the acceleration always is pointed towards the origin. Does the place of the sun even have any relevance? Considering that no masses and gravitation is taken account for.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3312/prove-that-vector-acceleration-always-points-towards-originSat, 15 Dec 2018 13:02:51 +0000Final charge on the capacitor and Charge on capacitor as a function time
http://physics.qandaexchange.com/?qa=3311/final-charge-the-capacitor-charge-capacitor-function-time
<p>In the given circuit diagram, initially charge on lower plate of capacitoris $Q_0$ = $CE/2$. Now switch is closed at t=0. Find<br>
1) Final charge on the capacitor<br>
2) Charge on capacitor asa a function time<br>
<img src="https://cdn.pbrd.co/images/HRxhcEy.jpg" alt=""></p>
<p>My try :<br>
<img src="https://cdn.pbrd.co/images/HRxhI0X.jpg" alt=""></p>
<p>In part b I tried to apply KVL but that would be quite lengthy.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3311/final-charge-the-capacitor-charge-capacitor-function-timeThu, 13 Dec 2018 12:16:20 +0000Check whether the followings are null, spacelike or timelike
http://physics.qandaexchange.com/?qa=3275/check-whether-the-followings-are-null-spacelike-or-timelike
<p>(a) <br>
$x^0=\int r^2+z^2d\tau$<br>
$x^1=\int r\sin\theta d\tau$<br>
$x^2=\int r\cos\theta d\tau$ <br>
$x^3=\int z d \tau$<br>
(b)<br>
$x^0= \sqrt3 ct$<br>
$x^1=ct$<br>
$x^2=ct_0 \sin(t/t_0)$<br>
$x^3=ct_0 \cos(t/t_0)$<br>
where $t_0$ is constant.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3275/check-whether-the-followings-are-null-spacelike-or-timelikeSun, 09 Dec 2018 11:01:29 +0000A projectile is ejected with vertical speed $v$ from the surface of a planet of mass $M$ and radius $R$
http://physics.qandaexchange.com/?qa=3295/projectile-ejected-with-vertical-speed-surface-planet-radius
<blockquote><p>A projectile is ejected with vertical speed $v$ from the surface of a planet of mass $M$ and radius $R$. Show that it comes to rest at a distance r from the centre of the planet where $$r=R/(1-Rv^2/2GM)$$ If $ v = c$ in the above example what will be the condition between M and R such that the planet acts like a Newtonian black hole.</p>
</blockquote>
<p>I have tried to solve it in the following way-</p>
<p>The height from the surface of the planet will be given by-<br>
$$y=vt-(1/2)gt^2$$<br>
when it comes to rest $v=0$ hence from the above equation we get<br>
$$r-R=(1/2)gt^2$$ <br>
From relation between g and G,<br>
$$g=Gm/r^2$$<br>
I have put all this values in $$y=vt-(1/2)gt^2$$<br>
but I am not getting the required answer. </p>
<p>Please tell me where did I made the mistake and correct me if going in the wrong direction. Thanking you. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3295/projectile-ejected-with-vertical-speed-surface-planet-radiusSun, 09 Dec 2018 11:01:24 +0000Rotation of L shaped rods after impulse
http://physics.qandaexchange.com/?qa=3304/rotation-of-l-shaped-rods-after-impulse
<blockquote><p>Two identical rod each of mass $m$ and length $\ell$ joined at one end and<br>
free to rotate about this end kept on smooth horizontal surface. An<br>
horizontal impulse $J$ is given to rod AB at centre and perpendicular to<br>
rod AB.<br>
<img src="https://cdn.pbrd.co/images/HQEyifd.png" alt=""><br>
<img src="https://cdn.pbrd.co/images/HQEyzJS.png" alt=""></p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HQEyNxw.jpg" alt=""><br>
I am just able to deal the angular velocity about centre of but unable to proceed for individual rods.</p>
<p>Unable to proceed for second part.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3304/rotation-of-l-shaped-rods-after-impulseFri, 07 Dec 2018 17:01:43 +0000System of two ideal Fermi gases in three dimensions.
http://physics.qandaexchange.com/?qa=3294/system-of-two-ideal-fermi-gases-in-three-dimensions
<blockquote><p>1) What is the pressure of a gas of free bosons in the limit of vanishing temperature, $T \rightarrow 0$?<br>
2)Argue that for $T \rightarrow 0$ an ideal Fermi gas will have non-vanishing pressure $p_0 > 0$. <br>
We will now use this fact to study a system of two ideal Fermi gases in three dimensions.<br>
A free sliding piston separates two compartments labeled 1 and 2 with volumes $V_1$ and $V_2$ respectively. An ideal Fermi gas with $N_1$ particles with spin 1/2 is placed in compartment 1 and an ideal Fermi gas with $N_2$ particles with spin 3/2 is placed in compartment 2.</p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HPrUESc.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HPrVk3f.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HPrVAq7.png" alt=""></p>
<p>As you notice, this problem has already been solved, but I do not understand the vast majority of it.</p>
<p>1) is OK.</p>
<p>2) I do not know how they got equation 15. It is stated to be a continuous approximation but not idea how to even start. </p>
<p>Actually, Griffiths has an interesting section in which treats the fermion distribution:</p>
<p><img src="https://cdn.pbrd.co/images/HPs57DJ.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HPs7CTH.jpg" alt=""></p>
<p>I know that the Fermi Dirac distribution (which is the one which interests us, since we are working with fermions) has a pretty well-known behavior as $T \rightarrow 0$ (please see figure 5.8 in Griffiths).</p>
<p>But I still do not know how to get it. </p>
<p>From here on I simply got lost. I mean, I have studied the free particle in QM but they go on with the density of states from EQ 16... I do not grasp it.</p>
<p>May you please shed some light on the provided solution (from EQ 15)?</p>
<p><strong>EDIT:</strong></p>
<p>Now I am stuck at EQ 19.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3294/system-of-two-ideal-fermi-gases-in-three-dimensionsThu, 29 Nov 2018 19:49:17 +0000The lean of a motorcyclist
http://physics.qandaexchange.com/?qa=3273/the-lean-of-a-motorcyclist
<blockquote><p>A motorcyclist takes a turn with a radius of $500 m$. Both the mass of the engine and the motorcyclist is $160 kg$ and their centre of mass lies $0, 5 m$ above the ground when it is perfectly vertical. His tangential speed is $60 m / s$ and its tangential acceleration is $4 m / s^2$.</p>
</blockquote>
<blockquote><p>a) Make a clear drawing in top view and back view with the<br>
speeds, accelerations and forces that affect the engine and motorcyclist.<br>
b) Calculate the speed and acceleration of the motorcyclist.<br>
c) Calculate the size of the components of all forces that act on<br>
the contact surface of the engine with the road.<br>
d) Calculate the angle at which the motorcyclist takes the turn (Hint: take its<br>
mass centre as the origin of the system).<br>
e) What is the apparent weight of the motorcyclist when he turns?* </p>
</blockquote>
<p>What I have done and what I do not know:</p>
<p>a)</p>
<p>I was not sure about what <strong>'top view and back view'</strong> meant so I did the following:</p>
<p>This is the diagram analysed from above: <a rel="nofollow" href="https://imgur.com/a/JtpNmfr">https://imgur.com/a/JtpNmfr</a></p>
<p>This is the diagram analysed from the ground: <a rel="nofollow" href="https://imgur.com/a/9AwOrWa">https://imgur.com/a/9AwOrWa</a></p>
<p>Note that the forces have just been drawn on the latest diagram because of clarity purposes. </p>
<p><strong>Is there anything missing on the diagrams?</strong></p>
<p>b) </p>
<p><a rel="nofollow" href="https://imgur.com/a/bbcrmPm">https://imgur.com/a/bbcrmPm</a></p>
<p><a rel="nofollow" href="https://imgur.com/a/0wPegtU">https://imgur.com/a/0wPegtU</a></p>
<p>NOTE: I assumed uniform circular motion. <strong>It makes sense for you what I did?</strong></p>
<p><strong>EDIT</strong></p>
<p>I was wrong assuming uniform circular motion because there is tangential acceleration; <strong>the motorcycle does not move with uniform speed.</strong></p>
<p>c) </p>
<p>Here we just have to use Newton's second law:</p>
<p>$$\sum F = Ma$$</p>
<p>As we are under uniform circular motion:</p>
<p>$$F = M a_c = M \frac{u^2}{OB}$$</p>
<p>However, we are not asked for calculating the centripetal force but the ones exerted on the road and the engine (always including the person's mass). These are:</p>
<p>The force exerted on the road by the engine:</p>
<p>$$F = Mg = -1569,6 N$$</p>
<p>The force exerted on the engine by the road:</p>
<p>This is simply the reaction force (Newton's third law):</p>
<p>N = 1569,6 N</p>
<p><strong>This seems too simple, am I missing something?</strong></p>
<p>d)</p>
<p>I do not know how to calculate the angle properly. Actually I got:</p>
<p>$$\alpha = \frac{ut}{OB}$$</p>
<p>But depends on time. <strong>Should I use kinematics to get the time and therefore the angle?</strong></p>
<p>e)</p>
<p><strong>Isn't the apparent weight just the normal force exerted on the motorcyclist? (i.e N = 1569,6 N)?</strong></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3273/the-lean-of-a-motorcyclistSat, 24 Nov 2018 17:45:39 +0000Questions from a jumping kangaroo
http://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangaroo
<p><img src="http://i68.tinypic.com/2lc8wnn.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangarooFri, 16 Nov 2018 15:38:20 +0000SHM with elastic collision.
http://physics.qandaexchange.com/?qa=3260/shm-with-elastic-collision
<blockquote><p>A ball suspended by a thread of length $l$ at the point $O$ on the wall, forming a small angle $\alpha$ with the vertical <a rel="nofollow" href="https://imgur.com/Ytg3szH">(Fig.)</a>. Then the thread with the ball deviated through a small angle $\beta$ $(\beta>\alpha)$ and set free. Assuming the collision of the ball with the wall to be perfectly elastic, find the oscillation period of such a pendulum.</p>
</blockquote>
<p><img src="https://i.imgur.com/g9zeh4t.png[/img]" alt=""></p>
<p>From <a rel="nofollow" href="https://imgur.com/a/7PTbF5W">fig</a> $2\theta$ part of SHM will be skipped over, $\cos\theta=\dfrac{\alpha}{\beta}\implies\cos2\theta=\dfrac{2\alpha^2-\beta^2}{\beta^2}$. So time it would be skpping over will be $t=\dfrac{1}{\omega}\cos^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)$. On substrating from total time $T=2\pi\sqrt{\dfrac{l}{g}}$ gives me $T'=2\sqrt{\dfrac{l}{g}}\bigg[\dfrac{\pi}{2}+\sin^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)\bigg]$. But in official answer argument of sin inverse is $\dfrac{\alpha}{\beta}$.</p>
<p>Please help.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3260/shm-with-elastic-collisionFri, 16 Nov 2018 13:31:41 +0000Diffusion with reflecting boundary
http://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary
<p><img src="https://cdn.pbrd.co/images/HNaMhdt.png" alt=""></p>
<p>What I have tried:</p>
<p><img src="https://cdn.pbrd.co/images/HNaMSPP.jpg" alt=""></p>
<p><img src="https://i.imgur.com/hls9bU5.png" alt=""></p>
<p>Note that my question is that how is it possible that I get $t = \infty$. It does not make sense for me (physically speaking). Where did I get wrong?</p>
<p>Please if there is information you require let me know. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundaryWed, 14 Nov 2018 20:07:35 +0000Angle made by the plane of hemisphere with inclined plane
http://physics.qandaexchange.com/?qa=3241/angle-made-by-the-plane-of-hemisphere-with-inclined-plane
<blockquote><p>A uniform thin hemispherical shell is kept static on an inclined plane of angle<br>
$\theta = 30$ as shown. If the surface of the inclined plane is sufficiently rough to prevent<br>
sliding then what is the angle $\alpha$ made by the plane of hemisphere with inclined plane .<br>
<img src="https://cdn.pbrd.co/images/HLyzG1g.jpg" alt=""></p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HLyzXc6.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3241/angle-made-by-the-plane-of-hemisphere-with-inclined-planeSun, 04 Nov 2018 04:53:51 +0000Radial distribution of particle separation in a liquid at small distances
http://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distances
<blockquote><p>Draw schematically the radial distribution function $g(r)$ for a Lennard-Jones fluid at low and high particle densities. Discuss in both cases the behavior at small $r$. At high densities $g(r)$ has some damped oscillations. Explain their origin and what would happen at long distances.</p>
</blockquote>
<p><strong>What I know:</strong></p>
<p>The radial distribution function $g(r)$ describes how density varies as a function of distance from a reference particle (<em>Wikipedia</em>).</p>
<p>Specifically in Lennard-Jones case, we have :</p>
<p><img src="https://i.imgur.com/Z41FKUL.png" alt="\[img\]https://i.imgur.com/Z41FKUL.png\[/img\]"></p>
<p>Here we can observe that if repulsion outweighs attraction, the curve will die out on the x axis. On the contrary, if it is the other way around, the curve will grow exponentially. For the sake of clarity, I would remark y = 0 as the Ideal Gas behaviour. </p>
<p>I have read that $g(r)$ vanishes at short distances, because the probability of finding two particles close to each other vanishes due to the repulsive part of the potential. At high densities $g(r)$ can show some damped oscillations: <br>
<img src="https://i.imgur.com/PqQPmcc.png" alt=""><br>
These oscillations express the preference of the particles to be found at specific distances from a reference particle at the origin. For instance in the LJ case, a first layer of particles will be localized closed to the minimum of LJ's potential, which is the origin of the first peak in $g(r)$.</p>
<p>This layer prevents other particles from getting close to it, which is what causes the first<br>
minimum in $g(r)$.</p>
<p><strong>What I do not know.</strong></p>
<p>1)The explanation: </p>
<p>'These oscillations express the preference of the particles to be found at specific distances from a reference particle at the origin'. </p>
<p><strong>How can particles have 'preference' to be at different distances?</strong> I think this is not a good physical explanation. I have been thinking about this and I would say that as there is a high density of particles at small distances, the repulsion force is exerted on these, which triggers such oscillations. Finally there is a point where they are so close to each other that the repulsion force provides them with an initial kinetic energy which will be equal to a final electrostatic potential energy of zero (selecting our zero of electrostatic potential energy at y=0) , and that is the moment when the curve dies out on the x axis. Do you agree with this explanation?**</p>
<p>2)Are we dealing with underdamped oscillations at high densities? Could you give an insight into these kind of oscillations (if what I have just said is not enough or incorrect)?</p>
<p>3)When we are dealing with long distances, would $g(r)$ tend to be one? What would we observe, the Ideal Gas behaviour? Why? Is it because there is no interaction between the particles?</p>
<p>As always, I am interested in explaining such phenomena from a physical point of view. </p>
<p><strong>ADDITIONAL INFORMATION RELATED TO THE RADIAL DISTRIBUTION:</strong></p>
<p>I came across the following slide:</p>
<p><img src="https://cdn.pbrd.co/images/HRnRJCC.png" alt=""></p>
<p>Which compares the structure factor $S(Q)$ with the radial distribution function $g(r)$. I know that the structure factor is a mathematical description of how a material <strong>scatters</strong> incident radiation, but <strong>how is the structure factor related to the distribution function (as this slide suggests)?</strong></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distancesThu, 01 Nov 2018 16:00:05 +0000Liquid in a capacitor as dielectric (IE Irodov 3.144)
http://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144
<blockquote><p>A parallel plate capacitor is located horizontally so that one of its plates is submerged into the liquid while the other is over its surface. The permittivity of the liquid is equal to $\epsilon$, its density is equal to $\rho$. To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface charge density $\sigma$?</p>
</blockquote>
<p>This is a type of controversial problem, as it contains a different answer by the different author, two different answers provided for this are </p>
<p> $$1) \ h=\dfrac{(\epsilon-1)\sigma^2}{2\epsilon_o\epsilon\rho g}$$ $$2) \ h=\dfrac{(\epsilon^2-1)\sigma^2}{2\epsilon_o\epsilon^2\rho g}$$</p>
<p><a rel="nofollow" href="https://ibb.co/fYmek0">Solution for 1</a> and <a rel="nofollow" href="https://youtu.be/oyfo-TJEG28?t=31">solution for 2</a></p>
<p>Please help!</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144Tue, 30 Oct 2018 01:37:42 +0000induced charge density at boundary surface
http://physics.qandaexchange.com/?qa=3216/induced-charge-density-at-boundary-surface
<blockquote><p>The space between the plates of a parallel plates capacitor is filled consecutively with two dielectrics layers $1$ and $2$ having the thickness $d_1$ and $d_2$ respectively and permittivities $\epsilon_1$ and $\epsilon_2$. The area of each plate is equal to $S$ find:<br>
1. The capacitance of the capacitor <br>
2. The density $\sigma'$ of the bound charges on the boundary plane if the voltage across the capacitor equals $V$ and the electric field is directed from layer $1$ to layer $2$.</p>
</blockquote>
<p>I managed to find $C_{eq}$ easily, for the second part I don't understand that word " <em>boundary plane</em>", I only know $\sigma'=\sigma\bigg(1-\dfrac{1}{\epsilon}\bigg)$ holds when there would have been just one plate, how to work on two such consecutive plates. Please help.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3216/induced-charge-density-at-boundary-surfaceSun, 28 Oct 2018 05:02:22 +0000Diffusion with an even number of reflecting boundaries
http://physics.qandaexchange.com/?qa=3211/diffusion-with-an-even-number-of-reflecting-boundaries
<p><img src="https://cdn.pbrd.co/images/HKoOk3T.png" alt=""></p>
<p>a)<br>
<img src="https://cdn.pbrd.co/images/HKoQ6sP.jpg" alt=""></p>
<p>This is the behaviour I expect from both concentrations at t= 0. Based on the theory of diffusion I would say that as t goes on, the Gaussian distributions will spread out and after a large number of collisions both concentrations will end up spreading out more or less evenly throughout the whole volume. Please let me know if this reasoning is wrong.</p>
<p>A doubt came to my mind as I imagine the same scenario if we had absorbing boundaries instead of reflecting ones. Therefore, does really matter the kind of boundary we are dealing with when we are trying to predict how the two concentrations (in this case) are going to spread out as time flows?</p>
<p>EDITED: </p>
<p>My initial idea about the identical scenario regarding either absorbing boundaries or reflecting ones was wrong. That is because after a long time in presence of absorbing boundaries, the concentration will end up being zero. </p>
<p>Related question: <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundary">http://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundary</a></p>
<p>b)<br>
<img src="https://cdn.pbrd.co/images/HKoXG2o.jpg" alt=""><br>
Based on what I have said before, I would expect the limiting case to be that both concentrations were equal to each other.</p>
<p>EDITED: </p>
<p>Actually I was wrong as there is only one distribution of concentration, which tends to be uniform after a long time (as explained at a)).</p>
<p><strong>I have been looking for a deeper explanation and I came up with the idea that at the boundaries the current is zero:</strong></p>
<p>$$j =-D\frac{\partial c(+-a/2,t)}{\partial x} = 0$$</p>
<p><strong>Which means that c(x,t) has vanishing derivatives. How could this fact help to explain the limiting behaviour of the concentration after a long time?</strong></p>
<p>c)<br>
This household potential:<br>
<img src="https://cdn.pbrd.co/images/HKp12yu.jpg" alt=""></p>
<p>As we are dealing with the distribution of particles in thermal equilibrium, the first thing that came to mind when I read potential was the Boltzmann equation:</p>
<p>$$n = ke^{-V(x)/KT}$$</p>
<p>EDITED: </p>
<p><strong>I am coming up with the Boltzmann distribution because it can tell us the distribution of molecules. If the potential energy is known as a function of distance, then the proportion of them at different distances is given by this law. And we do know the potential, which is the square-well. So in the equilibrium concentration I should be able to check whether with the square-well potential we get:<br>
- When $|x| \le \frac{a}{2}$ we get a constant (the value of the concentration in equilibrium between the boundaries).<br>
- When $|x| > \frac{a}{2}$ we get zero (outside the barriers).<br>
But how can I check this?</strong></p>
<p>d) <br>
<strong>I came across an explanation. It is not detailed and I barely understand it. Please explain it if you see that it makes sense:</strong> (I do not know why pasteboard did not work. I had to use Imgur):</p>
<p><a rel="nofollow" href="https://imgur.com/a/8TLgeRe">https://imgur.com/a/8TLgeRe</a></p>
<p><a rel="nofollow" href="https://imgur.com/a/unkkyyQ">https://imgur.com/a/unkkyyQ</a> </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3211/diffusion-with-an-even-number-of-reflecting-boundariesSat, 27 Oct 2018 15:15:35 +0000Charges induced on conducting plates.
http://physics.qandaexchange.com/?qa=3210/charges-induced-on-conducting-plates
<blockquote><p>Two infinite conducting plates $1$ and $2$ are separated by a distance $l$. A point charge $q$ is located between the plates at a distance $x$ from the plate $1$. Find the charges induced on each plate.</p>
</blockquote>
<p>Working on author's instruction, assuming $q$ to be spread uniformly on the plane through $q$ and parallel to plates, makes it easier to calculate electric field strength. Then to bring these distance into the picture we should work on potentials (as $E$ is known). But I don't know about potentials, should I take the potential difference between conductor to be zero, please explain everything in greater details.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3210/charges-induced-on-conducting-platesSat, 27 Oct 2018 05:01:02 +0000Relation between flux through lateral surface of a cylinder and flat parts.
http://physics.qandaexchange.com/?qa=3192/relation-between-flux-through-lateral-surface-cylinder-parts
<p>On solving IE Irodov 3.21, </p>
<blockquote><p>A ball of radius $R$ is uniformly charged with the volume density $\rho$. Find the flux of the electric field strength vector across the ball's section formed by the plane located at a distance $r_{0} < R$ from the center of the ball.</p>
</blockquote>
<p>From calculus obtained result is $$\boxed{\phi_1=\dfrac{\pi\rho r_o(R^2-r_o^2)}{3\epsilon_o}}$$</p>
<p>But if consider that flat surface as a flat surface of a cylinder then radius and height of such a cylinder will be $\sqrt{R^2-r_o^2}$ and $2r_o$ respectively, then from Gaus' Law flux through <strong>whole of cylinder</strong> is given by $$\boxed{\phi_2=\dfrac{\pi\rho (2r_o)(R^2-r_o^2)}{\epsilon_o}}$$</p>
<p>This is very similar to flux $\phi_1$ obtained for flat surface. More precisely $$\phi_1=\dfrac{\phi_2}{6}$$</p>
<p>So can we prove that flux through flat part of the cylinder is one-third of the total, and that through curved part is two-thirds of the total, in general?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3192/relation-between-flux-through-lateral-surface-cylinder-partsMon, 22 Oct 2018 12:21:17 +0000Oscillations of a rotating mass on a spring
http://physics.qandaexchange.com/?qa=3191/oscillations-of-a-rotating-mass-on-a-spring
<blockquote><p>A spring of constant $k$ is wrapped around a long rod. One end of the rod and spring are attached to a motor that rotates the rod with constant angular velocity $\Omega$. The other end of the spring is attached to a mass $m$. This mass slides along the rod as the spring expands and contracts. There is no friction or gravity in the problem. The equilibrium length of the spring is $r_{0}.$</p>
</blockquote>
<p><img src="http://i68.tinypic.com/2crm0pc.png" alt=""></p>
<blockquote><p>What is the oscillation frequency of the mass for small values of $\Omega$? What happens when $\Omega$ is large? Is the motion still simple harmonic motion?</p>
</blockquote>
<p>Since there is no gravity, the centrifugal force must equal the spring force so</p>
<p>$$m\Omega^{2}r=k(r-r_{0})$$</p>
<p>which leads to a constant $r$ since $\Omega$ is constant. So I am missing something basic here.</p>
<p>Maybe I should use energy conservation?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3191/oscillations-of-a-rotating-mass-on-a-springSun, 21 Oct 2018 20:39:57 +0000Deducing the mass of the Sirius system - Feynman exercises 3.17
http://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17
<p><img src="https://cdn.pbrd.co/images/HJlk9yQ.jpg" alt=""></p>
<p>I do not even know how to try to deduce the mass M of the Sirius system in terms of that of the sun, as I do not not how to interpret the given data. </p>
<p>Could you give me a hint? </p>
<p><strong>EDITED AT THIS POINT</strong></p>
<p><strong>1) Estimating the distance to the Sirius binary system.</strong></p>
<p>I calculated the hypotenuse of the triangle $S E_1 SS$ (Sun - position of the Earth located at the top of a vertical circular orbit - Sirius System):</p>
<p>$$ \theta = 0.378^{\circ} arc \times \frac{1}{3600} \times \frac{2 \pi}{360} = 1.83 \times 10^{-6}arc sec $$</p>
<p>$$D = \frac{D'}{sin(\theta)} = 8.16\times 10^{16} m$$</p>
<p>Where:</p>
<p>D = the distance from the Earth to the Sirius binary system.</p>
<p>D' = Distance from the Sirius system to the Earth ($15 \times 10^{10} m$).</p>
<p>Here's an illustration of what I have done:</p>
<p><img src="https://cdn.pbrd.co/images/HJtrflM.png" alt=""></p>
<p><strong>2) Measuring the semi-major axis of the ellipse and estimating the period of the orbit from the figure.</strong></p>
<p>To compute the semi-major axis I have assumed the major axis is under the scale [0'',12'']. </p>
<p>As Sammy Gerbil said :'To measure the semi-major axis from the figure, magnify the diagram to fill your screen, place the edge of a sheet of paper along the diagonal line marked on the ellipse, mark off the ends of this line on the paper, then transfer the paper edge to the scale and read off the length in arc seconds. I get 13.8" for the major axis length, so the semi-major axis is a=6.9". The calculation is then:'</p>
<p>$$a=8.16\times 10^{16}m\times \frac{6.9}{3600}\times \frac{2\pi}{360} = 2.73 \times 10^{12}m$$</p>
<p>Now let's estimate the period of the orbit.</p>
<p>Is this the right thought? </p>
<p>$$\frac{dA}{dt} = SM \times SM \times \frac{d \theta}{2dt}$$</p>
<p>It will not be accurate as 'the two sides of most of the triangles are different'.</p>
<p>$A_1$:</p>
<p><img src="https://cdn.pbrd.co/images/HK5k6ab.jpg" alt=""></p>
<p>On process...</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17Sat, 20 Oct 2018 15:50:53 +0000Find the angle between the $x$-axis and a vector
http://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vector
<blockquote><p>The $x$ component of vector $A$ is 25.0 m and the $y$ component is 40.0 m. <br>
(a) What is the magnitude of $A$? <br>
(b) What is the angle between the direction of $A$ and the positive direction of x?</p>
</blockquote>
<p>For (b) I tried using the formula $\tan \theta = \frac{a_y}{a_x} = \frac{40}{-25} = -1.6$, thus $\arctan(-1.6)=58$ degrees which does not match the answer key: $122$ degrees.</p>
<p>Any help is appreciated.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vectorSat, 20 Oct 2018 11:59:19 +0000Onset of bouncing for rolling hoop with off-centre mass (Irodov ex 1.265)
http://physics.qandaexchange.com/?qa=3173/onset-bouncing-for-rolling-hoop-with-off-centre-mass-irodov
<blockquote><p>A small body $A$ is fixed to the inside of a thin rigid hoop of radius $R$ and the mass equal to that of the body $A$. The hoop rolls without slipping over a horizontal plane, at the moments when the body $A$ gets into lower position, the center of the hoop moves with velocity $v_0$ At what values of $v_0$ will the hoop move without bouncing.<img src="https://cdn.pbrd.co/images/HJaN8cM.png" alt=""></p>
</blockquote>
<p>It's very intuitive that bouncing can happen only at the top point.</p>
<p>Let there be normal $N$ at the top between $A$ and hoop and hoop rotating with $\omega$, for centripetal force and condition that $N$ lifts up $mg$ of hoop$$N+mg=m\omega^2R\\ N=mg\\ \implies 2mg=m \omega^2R\implies \omega^2=\dfrac{2g}{R}$$</p>
<p>Now I need to conserve energy, the best option for most cases is from CM but conserving it here from CM makes it tedious, so from where and how should I conserve it to get it easily.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3173/onset-bouncing-for-rolling-hoop-with-off-centre-mass-irodovFri, 19 Oct 2018 12:56:10 +0000Normal reaction for particle at rest on a sphere
http://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphere
<p><img src="https://cdn.pbrd.co/images/HIVCKHf.png" alt=""></p>
<p>The answer is C. </p>
<p>It is not difficult to understand why force $F$ decreases as the small metal ball is raised to the top of the sphere. The normal reaction $R$ from the hemisphere bears a greater share of the weight of the metal ball. </p>
<p>But why is the normal reaction $R$ the same for all positions of the ball on the sphere? Is there an intuitive explanation?</p>
<p>See discussion in <a rel="nofollow" href="https://chat.stackexchange.com/transcript/message/47213469#47213469">Problem Solving Strategies chatroom</a> on Physics Stack Exchange.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphereWed, 17 Oct 2018 22:19:45 +0000The length of the spring
http://physics.qandaexchange.com/?qa=3166/the-length-of-the-spring
<blockquote><p>Turns of a uniform spring of relaxed length l = 1.00 m and force constant<br>
k = 500 N/m almost touch each other. A light glue is applied evenly<br>
between every adjacent turn. Breaking strength of the glue is $F_b = 100<br>
N$. The spring is placed on a frictionless horizontal floor and pulled from<br>
one of its ends. If the pulling force is gradually increased to a value F = <br>
200 N. how much will the length of the spring become?</p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HIzqCh6.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3166/the-length-of-the-springMon, 15 Oct 2018 13:43:26 +0000The Random Walk
http://physics.qandaexchange.com/?qa=3164/the-random-walk
<p>Consider a one dimensional random walk performing discrete jumps of length $a$ at each time step.</p>
<p>a) Calculate $P(p,N)$ the probability that the walk of $N$ steps performs $p$ steps to the right.</p>
<p><strong>I do not know if the binomial probability approach can be applied to this problem. If that is the case, why?</strong></p>
<p><img src="https://cdn.pbrd.co/images/HIksJH1.jpg" alt=""></p>
<p>The information I used:</p>
<p><img src="https://cdn.pbrd.co/images/HIktjBc.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HIktxSa.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HIktJQ3.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HIku2ST.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3164/the-random-walkSat, 13 Oct 2018 23:42:35 +0000Find angular acceleration of disc and tension in the string
http://physics.qandaexchange.com/?qa=3155/find-angular-acceleration-of-disc-and-tension-in-the-string
<blockquote><p>A uniform disc of mass m and radius R is kept on frictionless horizontal table. Two particles of mass m are connected to disc by two identical light inextensible threads as shown in figure. The particles are given velocity $v_0$ perpendicular to the length of strings. Then find angular acceleration of disc and tension in the string .<br>
<img src="https://cdn.pbrd.co/images/HHxIiMM.jpg" alt=""></p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HHxIzPN.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3155/find-angular-acceleration-of-disc-and-tension-in-the-stringMon, 08 Oct 2018 19:35:17 +0000Maximum height after collision in a bowl
http://physics.qandaexchange.com/?qa=3146/maximum-height-after-collision-in-a-bowl
<p>Consider two masses $m, M$ that fall from the top of a hemispherical bowl (from rest) and slide down without any friction.</p>
<p>We want to find how high each of the masses will go after the collision assuming the collision is elastic.</p>
<p><img src="http://i63.tinypic.com/23vnrxv.jpg" alt=""></p>
<p><img src="http://i68.tinypic.com/5v20sn.png" alt=""></p>
<p><img src="http://i68.tinypic.com/28w1937.png" alt=""></p>
<p><img src="http://i65.tinypic.com/b49hdi.png" alt=""></p>
<p>This is how far I can go. It seems to me there are too many unknowns and I am missing some key simplification.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3146/maximum-height-after-collision-in-a-bowlMon, 08 Oct 2018 01:15:46 +0000Thevenin's theorem on Capacitor.
http://physics.qandaexchange.com/?qa=3144/thevenins-theorem-on-capacitor
<p>The time constant of the circuit shown is:<br>
<img src="https://cdn.pbrd.co/images/HHlOycn.png" alt=""><br>
I found $R_{th}=\dfrac{3R}{5}$ by following method :</p>
<p>We short-circuit the batteries to calculate Thevenin's equivalent resistance across $C$ hence $2R\parallel 2R=R$ which is parallel with $\dfrac{R}{2}+R=\dfrac{3R}{2}\implies R_{th}=R\parallel \dfrac{3R}{2}=\dfrac{3R}{5}$ <br>
So time constant $\tau=\dfrac{3RC}{5}$</p>
<p>But $\tau=\dfrac{RC}{2}$, given. Where I went wrong? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3144/thevenins-theorem-on-capacitorSun, 07 Oct 2018 13:26:10 +0000Maximum energy stored in a spring-block system
http://physics.qandaexchange.com/?qa=3124/maximum-energy-stored-in-a-spring-block-system
<blockquote><p>A block of mass $m$ is attached with an ideal spring of spring constant $k$ is kept on a smooth horizontal surface. Now the free end the spring is pulled with a constant velocity $u$ horizontally. The maximum energy stored in the spring and block system during subsequent motion is?</p>
</blockquote>
<p>I think,<br>
Spring will keep on extending till it gain velocity $u$ to stop free end's expansion, then spring is elongated it will apply force on mass $m$ till it again gain its natural length, by now maximum work has been done, so if I calculate $v_m$ then $E=\frac{1}{2}mv^2$, so how to calculate $v_m$?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3124/maximum-energy-stored-in-a-spring-block-systemSat, 06 Oct 2018 05:01:17 +0000Finding path for which line integral of force is zero.
http://physics.qandaexchange.com/?qa=3123/finding-path-for-which-line-integral-of-force-is-zero
<blockquote><p>A particle is constrained to move from initial point $O$ to final point $C$ along three different smooth horizontal tracks namely $OBC, OPC, OAC$. If the particle moves under the influence of an external force $F$ such that the initial and final speeds are same then:<br>
1. There necessarily exists a path along which line integral of force $F$ is zero.<br>
2. $F$ is conservative<br>
3. $F$ cannot be conservative<br>
4. There is no closed path along which line integral of force $F$ is zero.</p>
</blockquote>
<p>Only correct answer is $(1)$.</p>
<p>What does this mean?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3123/finding-path-for-which-line-integral-of-force-is-zeroSat, 06 Oct 2018 03:17:35 +0000Average of the internal energy of a system
http://physics.qandaexchange.com/?qa=3122/average-of-the-internal-energy-of-a-system
<p><img src="https://cdn.pbrd.co/images/HH4nYLV.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HH4odz4.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HH4oOvr.jpg" alt=""></p>
<p>I am struggling to get the right value for $Z_1$, which is $\beta^{\frac{-5}{2}}$ multiplied by constants. Once you have the value for the one particle partition function it is just about applying the definition of average energy. </p>
<p>$$\langle E \rangle = -\frac{ \partial \ln(Z_1)}{\partial \beta}$$</p>
<p>I got :</p>
<p>$$ \langle E \rangle = \frac{9}{4\beta}$$</p>
<p>The right answer is though:</p>
<p>$$\langle E \rangle = \frac{5}{2\beta}$$</p>
<p>NOTE: I know this is more a mathematical issue and I have already asked for this question in MSE, but I could not get the right value for < E > yet. If you want to have a look:</p>
<p><a rel="nofollow" href="https://math.stackexchange.com/questions/2941009/solving-a-partition-function-in-polar-coordinates">https://math.stackexchange.com/questions/2941009/solving-a-partition-function-in-polar-coordinates</a></p>
<p>More information about the partition function and more definitions if needed be. Please let me know if you need information that is not stated:</p>
<p>$$\beta = \frac{1}{k_B T}$$</p>
<p><img src="https://cdn.pbrd.co/images/HH4rh4c.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HH4rrF1L.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HH4rGeA.png" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3122/average-of-the-internal-energy-of-a-systemFri, 05 Oct 2018 17:17:37 +0000The ratio of temperatures of two sides of a disk near a concave mirror
http://physics.qandaexchange.com/?qa=3117/the-ratio-of-temperatures-two-sides-disk-near-concave-mirror
<blockquote><p>A wide homogeneous beam of light falls on a concave spherical mirror of<br>
radius R parallel to the optical axis. A small opaque disc of radius r (r<br>
<< R) made of a perfectly heat insulating material is placed at a distance<br>
R/4 from the pole of the mirror perpendicular to the optical axis. In<br>
steady state, both the surfaces of the disc acquire different temperatures<br>
slightly higher than the surroundings. Find the ratio $\delta T_1 /\delta T_2$, where<br>
$\delta T_1$. and $\delta T_2$; are the temperature differences of the left and the right<br>
surfaces of the disc and the surroundings.<br>
<img src="https://cdn.pbrd.co/images/HGqS17f.jpg" alt=""></p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HGqSh7X.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3117/the-ratio-of-temperatures-two-sides-disk-near-concave-mirrorMon, 01 Oct 2018 12:24:00 +0000Is centripetal force not always true for Planets?
http://physics.qandaexchange.com/?qa=3114/is-centripetal-force-not-always-true-for-planets
<p><img src="https://cdn.pbrd.co/images/HGi2DtH.png" alt=""></p>
<p><em>Source :</em> IE Irodov: 1.207.</p>
<p>I used, force by gravitational pull of Sun should provide necessary centripetal force at one extreme $r_1$ where velocity is $v_1$ ( I took that moment as circle as $r_1\perp v_1$) ,$$\implies\dfrac{GM_sm}{r_1^2}=\dfrac{mv_1^2}{r_1}\implies v_1=\sqrt{\dfrac{GM_s}{r_1}}$$</p>
<p>But when I use energy conservation then, $$v_1=\sqrt{\dfrac{2GM_sr_2}{r_1(r_1+r_2)}}$$</p>
<p>Second expression gives me the correct answer, also if I put $r_1=r_2$ second expression reduces to first one, which says to me that we apply centripetal force method only when we have complete circle, not that instantaneous circles, why?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3114/is-centripetal-force-not-always-true-for-planetsSun, 30 Sep 2018 11:04:59 +0000Solving the diffusion equation with an absorbing boundary
http://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundary
<p><img src="https://cdn.pbrd.co/images/HGatckL.png" alt=""></p>
<p>There is a one-dimensional diffusion process in which particles start running at $t = 0$ and from $x_o > 0$.</p>
<p><strong>When particles reach x = 0 they are removed from the system, thus the total concentration is not conserved anymore.</strong></p>
<p>I have to solve the diffusion equation, which is the following partial differential equation:</p>
<p>$$\frac{\partial P (R, t)}{\partial t} = D\triangledown^2P(R,t) $$</p>
<p>Where $P(R, t)$ is the probability that the particles arrive at R at time t.</p>
<p>We have the initial conditions:</p>
<p> $$c(x,0) = N\delta(x - x_o)$$</p>
<p> $$c(0, t) = 0$$</p>
<p>I have been doing some research in how to do so and I came across with a method which is based on a particular Gaussian function:</p>
<p>$$G (R, t) = (\frac{1}{4\pi Dt})^{\frac{d}{2}} e^{\frac{(R-R_0)^2}{4Dt}}$$</p>
<p>Where d is the dimensionality of the system.</p>
<p>But the issue here is that we are working with an 'absorbing boundary' that makes the condition $c(0, t) = 0$ useless because we work from $x_o > 0$. </p>
<p>Then how could I solve the probability (i.e this differential equation)? We have been suggested the method of images, but not sure how it works with differential equations.</p>
<p>NOTE: I know this is more a mathematical issue (solving a differential EQ.) but thought that as it is a mechanics problem it could be useful asking here. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundarySat, 29 Sep 2018 18:43:19 +0000Angle of pendulum in accelerated incline
http://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-incline
<p>Consider the following system consisting of a box sliding down a plane. The coefficient of friction between the plane and the box is $\mu$. A pendulum is attached to the top of the box as shown.</p>
<p><img src="http://i65.tinypic.com/1z37azr.png" alt=""></p>
<p>The acceleration of the box+pendulum is $a=g(\sin\theta-\mu\cos\theta)$ I believe.<br>
In a non-inertial frame attached to the box, the free-body diagram for the pendulum is</p>
<p><img src="http://i63.tinypic.com/2eyyx3s.png" alt=""></p>
<p>My goal is to find the angle $\phi$ from the equilibrium of these 3 forces. I have to pick x and y axes to decompose these forces. If I pick the x axis along the fictitious force and the y perpendicular to it I get (I think) </p>
<p>$$m_P a+T\sin\phi=m_P g\sin\theta$$ </p>
<p>and </p>
<p>$$T\cos\phi=m_{P}g\cos\theta$$</p>
<p> which I can then solve for $\phi$ :</p>
<p>$$\tan\phi=\frac{g\sin\theta-a}{g\cos\theta}=\mu$$</p>
<p>Is this right?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-inclineFri, 28 Sep 2018 04:00:23 +0000Minimum distance from the mirror should the boy be to see his full image
http://physics.qandaexchange.com/?qa=3099/minimum-distance-from-the-mirror-should-the-boy-see-full-image
<blockquote><p>A plane mirror with its bottom edge on the floor is tilted at an angle $\theta$<br>
to the vertical (see figure). A boy whose eyes are at a height h above the<br>
floor is standing in front of the mirror. At what minimum distance from<br>
the mirror should the boy be to see his full image in the mirror?<br>
<img src="https://cdn.pbrd.co/images/HFvJzFb.jpg" alt=""></p>
</blockquote>
<p><img src="https://cdn.pbrd.co/images/HFvJOfm.jpg" alt=""><br>
To get the minimum d the rays should go parallel to floor and then get reflected to the boy eyes<br>
Answer given = $(h\sec\theta)/2$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3099/minimum-distance-from-the-mirror-should-the-boy-see-full-imageTue, 25 Sep 2018 10:59:43 +0000Lengths of vertical cables and maximum tension - Feynman exercises 2.35
http://physics.qandaexchange.com/?qa=3088/lengths-vertical-cables-maximum-tension-feynman-exercises
<p><img src="https://cdn.pbrd.co/images/HFc4uGb.png" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HFc0vgU.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HFojRUM.jpg" alt=""></p>
<p><strong>a) Find the proper lengths for the remaining vertical cables A and B.</strong></p>
<p>I focused on the left hand side of the diagram and drew the right triangle BEC. Once here I drew the bisector of the right angle located at C, originating the right triangle CED. Now to get the angle $\theta$ we just need to know that the three angles of a triangle sum 180 degrees. As you can see I applied that doing:</p>
<p>angle E + angle CDE + angle DCE = 180 (where angle E = $\theta$ )</p>
<p>So now it is just about solving for $\theta$, which I got is 45 degrees. This is not correct so I am wondering what I did wrong here. Note after I applied tan $\theta$ in order to get the length BC which, as expected, did not make sense.</p>
<p>I am sure my mistake has to be on calculating $\theta$ but I do not see where is the flaw exactly.<br>
It might be on calculating the angle DCE but it has to be 45 degrees because is the bisection of a right angle.</p>
<p>To find the distance GC (the second largest cable, the one related to letter A in the original exercise) I could not find the proper method.</p>
<p>Given answers A = 5 m; B = 11 m.</p>
<p><strong>b) Find maximum tension in the two longitudinal cables</strong></p>
<p>$$\sum F_y = mg $$ </p>
<p>$$T sin 45 - mg = 0$$</p>
<p>$$T = 6.66 x 10^5 N$$</p>
<p>But the given answer is T = 34 x 10^3 kg-wt</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3088/lengths-vertical-cables-maximum-tension-feynman-exercisesSun, 23 Sep 2018 09:19:27 +0000