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Distance travelled by block in rotational motion

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The disk shown has weight 10 kg. Cylinder A and block B are attached to cords that are wrapped on the pulley as shown. The coefficient of kinetic friction between block B and the surface is 0.25. Knowing that the system is released from rest in the position shown, determine the total distance that block B moves before coming to rest.

I tried as
Let the tension in string attached to block B be T$_1$
$T_1 -f=5a_1$
torque equation $T_1/10=(10/100)\alpha$
$10\alpha = a_1$

For tensionin string attached to blockA $T_2$
force equation : 100$-T_2=10 a_2$
Torque equation : $T_2=(10/25)(\alpha )$
$\alpha =20a_2$

But solving this I am not getting the answer

asked Feb 3, 2017 in Physics Problems by koolman (4,286 points)
edited Jul 20, 2018 by sammy gerbil

1 Answer

2 votes
 
Best answer

Revised Answer

Such problems are usually solved most easily using conservation of energy.

Before A hits the ground both strings remain taut. The distance B moves is half the distance A moves, and the speed of B is half that of A. The speeds of A and B at any time can be related to the angular velocity of the pulley, assuming there is no slipping between strings and pulley.

So you can write that at any instant before A hits the ground :
(PE lost by A) = (KE gained by A, B and pulley) + (work done against friction by B).

When block A hits the ground its KE is dissipated as heat and sound so it is removed from the system. The KE which remains in the system equals the further work done against friction by block B, because the system is finally at rest and the only way that the KE in the system can be dissipated is by friction at B. This allows you to find the distance B moved after A hit the ground.

Add the distance moved by B before A hit the ground (0.5m) to the distance it moved after.

answered Feb 3, 2017 by sammy gerbil (28,746 points)
selected Feb 4, 2017 by koolman
Why we will not consider when $I = (1/2)MR^2 _B$ and $\omega = v_B / R_B$
$\omega=v_A/R_A=v_B/R_B$ is correct and consistent. However, the MI of the pulley cannot be both $\frac12MR_A^2$ and $\frac12MR_B^2$. These are not the same because $R_A \ne R_B$.

The pulley clearly extends to $R_A=20cm$. The question is, is the pulley a single uniform disk of radius $20cm$ or a composite of 2 disks of radii $10cm$ and $20cm$? If the latter, what is the mass of each disk?

The question does not provide enough information. The simplest assumption is that the pulley is a single disk which is uniform with radius $20cm$.
And Using the equation given by me above and KE of pulley as 10$v^2$ , I am getting the $v^2 = (375)/(130)$ . What mistake I have done .
I get the same answer. What answer have you been given? Sorry, you already said it is $\sqrt{\frac{150}{13}}$.

This is the same as your answer : $v_A=2v=2\sqrt{\frac{375}{130}}=\sqrt{\frac{150}{13}}$.
Thanks a lot
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