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The lean of a motorcyclist

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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 $4 m / s^2$.

a) Make a clear drawing in top view and back view with the
speeds, accelerations and forces that affect the engine and motorcyclist.
b) Calculate the speed and acceleration of the motorcyclist.
c) Calculate the size of the components of all forces that act on
the contact surface of the engine with the road.
d) Calculate the angle at which the motorcyclist takes the turn (Hint: take its
mass centre as the origin of the system).
e) What is the apparent weight of the motorcyclist when he turns?*

What I have done and what I do not know:

a)

I was not sure about what 'top view and back view' meant so I did the following:

This is the diagram analysed from above: https://imgur.com/a/JtpNmfr

This is the diagram analysed from the ground: https://imgur.com/a/9AwOrWa

Note that the forces have just been drawn on the latest diagram because of clarity purposes.

Is there anything missing on the diagrams?

b)

https://imgur.com/a/bbcrmPm

https://imgur.com/a/0wPegtU

NOTE: I assumed uniform circular motion. It makes sense for you what I did?

EDIT

I was wrong assuming uniform circular motion because there is tangential acceleration; the motorcycle does not move with uniform speed.

c)

Here we just have to use Newton's second law:

$$\sum F = Ma$$

As we are under uniform circular motion:

$$F = M a_c = M \frac{u^2}{OB}$$

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:

The force exerted on the road by the engine:

$$F = Mg = -1569,6 N$$

The force exerted on the engine by the road:

This is simply the reaction force (Newton's third law):

N = 1569,6 N

This seems too simple, am I missing something?

d)

I do not know how to calculate the angle properly. Actually I got:

$$\alpha = \frac{ut}{OB}$$

But depends on time. Should I use kinematics to get the time and therefore the angle?

e)

Isn't the apparent weight just the normal force exerted on the motorcyclist? (i.e N = 1569,6 N)?

asked Nov 24, 2018 in Physics Problems by Jorge Daniel (606 points)
edited Nov 30, 2018 by Jorge Daniel
I know that the English used on the exercise is not the best (it has been translated using Google Translator). Please let me know if something is not clear.

Thank you for the link. I will update the question once I have checked it.
OK The video answers explicitly a) and d).

Actually my approach is totally wrong, as I worked using just uniform circular motion.

Some remarks:

a) Does  top view and back view mean before it bends (perfectly vertical) and once it has bent respectively?

b) Aren't velocity and acceleration already given on the initial text?

c) Is it just about calculating the weight and its normal reaction right?

e) Just the normal, right?

Thanks
a) Top view means looking down from above,  so that you see the motorcycle moving in a circle. With this view you can draw vectors representing velocity, tangential acceleration and radial (centripetal) acceleration, and the radius of the turn.

Back view means looking at the motorcycle from behind, so that you see it tilted towards the centre of the circle. With this view you can show the angle of tilt, and the normal, centripetal and gravity forces.

b) Tangential acceleration is given, centripetal (radial) acceleration is not. Total acceleration is the vector sum of tangential and radial accelerations.

c) There is normal contact reaction (vertical), but also frictional contact forces which are tangential and radial (both horizontal). So there are 3 components of the contact force to be calculated.

e) Yes, apparent weight is just the normal component of the contact force.
Please if there is something you do not agree with, let me know.

Remaining doubts:

b) OK, I realised that the calculated centripetal acceleration was eventually OK. This: https://imgur.com/a/0wPegtU is simply:

$$a_c = \frac{u}{OB} = 7,2 m/s^2$$

Where $u$ is the tangential speed and $OB$ the radius.

Then, to get the total acceleration we just have to add the tangential acceleration:

$$a = a_c + a_t = 11,2 m/s^2$$.

c) I started from scratch here (based on this diagram: https://imgur.com/a/PrzlEsV), in order to see if I could solve the problem based on conservation of angular momentum.

In this system the AM is conserved (based on an axis passing through the centre of mass of the system and parallel to tangential speed) **if we neglect friction**, because the net external torque acting on the system is constant. This is because, about the axis, the angular acceleration is zero:

$$\alpha = \frac{d \omega}{dt} = 0$$

(we can also argue that Newton's second law for rotation is valid for any moving object if its rotation is about an axis passing through its centre of mass).

Based on $\sum\tau = 0$ and $\tau = r F$:

$$F_N d sin\theta -F_{fr} d cos \theta = 0$$

Then:

$$tan \alpha = \frac {F{fr}}{F_N}$$

Once here we can calculate the angle, which yields 36.28 º.

The magnitude of frictional force, normal force and gravitational force are:

$$F_N = Mgcos \theta = 1265.31 N$$

$$F_{fr} = \frac{M v^2 sin \theta}{r} = 681,67 N$$

$$W = -Mg = -1569.6 N$$

NOTE that d is the distance from the origin to the CM and I changed the notation of the angle (because the angular acceleration is defined to be $\alpha$).
However, I think I am missing something calculating the forces because they do not add up exactly zero. Do you see why am I wrong?
e) Then the apparent weight it is just $F_N = Mgcos \theta = 1265.31 N$, Yes?
OK I’ve just realised AM is not conserved due to friction... I guess we can neglect friction and apply the stated reasoning. However, I do not get $\sum F=0$ either...
(b) As I wrote in the last comment, total acceleration requires **vector addition** because acceleration is a vector.

(c) Why should the total force on the motorcycle be zero? It is accelerating both radially and tangentially, so the resultant force on it is not zero.

(e) There is no vertical acceleration so the resultant force in the vertical direction must be zero. The only vertical forces are normal reaction and weight.

You cannot neglect friction because without it there would be no centripetal acceleration and no tangential acceleration. Friction is the only force causing centripetal acceleration. So if your radial friction force does not equal the centripetal force you've done something wrong.
b) My bad, it would be:

$$a_{tot} = \sqrt{ a_t^2 + a_c^2 } = 8,24 m/s^2$$

c) I thought $\sum F=0$ because the motorbike has to be in equilibrium when leaning. Otherwise it would either fall or go up. By the way, do you agree with the calculation of the three forces?

e) Sorry professor but I have to say that I do not agree. I think there is vertical acceleration: gravity acts on the CM of the system (motorbike + person). Please tell me what I am missing here.

Assuming that there is conservation of AM I got:

$$tan \alpha = \frac {F{fr}}{F_N}$$

But I explicitly included the friction force, which means AM is not conserved.

Therefore the first thinking is: as the torque of the system is not zero, better if we do not talk about AM in this system.

But after I thought, hey, why don't we start like this? : the weight pointing downwards (as previously stated) and the normal force pointing at the CM of the system (as explained in the video you shared). We know that the horizontal component of the normal reaction is equal to the friction force, but in this scenario I think we can talk about conservation of AM because the normal force passes through the CM, meaning that:

$$\frac{dL}{dt} = \sum \tau = 0$$

Do you agree?
Basically what I am concerned about is how to get:

$$tan \alpha = \frac {F{fr}}{F_N}$$

Based on AM conservation and without using $F_{fr}$.
(c) The resultant force on the motorcycle is zero if you use a frame of reference which rotates with the motorcycle. For this you need to include the centrifugal force, acting on the centre of mass.

No I do not agree with your calculation of normal force $F_N$ and (radial) friction force $F_f$. You should have $F_N=W$ because these are the only vertical forces - there is no $cos\theta$ factor. Also $F_f=Ma_c$ because this is the only radial force - there is no $\sin\theta$ factor.

There is also a tangential frictional force $Ma_t$ which you have neglected.

$\tan\theta=\frac{F_f}{F_N}$ follows from the fact that $\theta$ is constant, so there is no moment about the COM of the motorcycle. The resultant or $F_N$ and  $F_f$ must pass through the COM. See diagram below.

Conservation of angular momentum (in the rear view) means that the resultant moment of forces on the motorcycle about the COM is zero. You have already made this calculation.
c) Regarding this comment:

'No I do not agree with your calculation of normal force $F_N$ and (radial) friction force $F_f$. You should have $F_N=W$ because these are the only vertical forces - there is no $cos \theta$ factor. Also $F_f=Ma_c$ because this is the only radial force - there is no $sin \theta$ factor.'

As the diagram states, $F$ is the sum of $N$ and $F_f$. I would say that for this to be true, we should have:

$$F = Fsin \theta + Fcos \theta$$

Having:

$$Fsin \theta = F_f = F_c$$

$$Fcos \theta = F_N = -W$$

Do you agree now?

By the way, What's $F$? On the video you shared, is treated as another reaction force (it is given the letter $R$). OK this is a silly question. R is just the tilted reaction, that is why the following equation holds:

$$Fcos \theta = F_N = -W$$

CALCULATIONS

$$F_N = Mg = 1569.6 N$$

$$f_r = \frac{M v^2}{r} = 1152 N$$

The tangential frictional force:

$$f_t = Ma_t = 640 N$$

Thus, the resultant force is the vector sum of the two frictional forces:

$$\sum F = 1317.84 N$$

Do you agree?

NOTE: I was wrong thinking that $\sum F=0$. However, the angular momentum is conserved due to $\sum \tau=0$
e) $F_N = 1569.6 N$
The diagram which I provided states that $F$ is the (vector) sum of $F_N$ and $F_f$. It is not *another* reaction force. Neglecting the tangential friction force (ie assuming tangential acceleration is zero) $F$ is the vector resultant (ie total) reaction force on the  bike from the road.

Your question does not ask you to calculate $F$ so it is not worthwhile to use it in calculations. It asks you for the values of the separate *components* of the reaction/contact forces.

Writing Newton's 2nd law vertically, horizontally (ie radially) and tangentially we have $$N-W=0$$ $$f_r=ma_c$$ $$f_t=ma_t$$ That is all you need to do.
Just for the sake of curiosity: the resultant force would simply be: $\sum F= f_r + f_t$.
Yes, assuming that is vector addition.

1 Answer

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Best answer

The back/front view should look like this :

The vertical forces $N$ and $W$ cancel out. The resultant force is a radial friction force $f$ acting towards the centre of the turning circle. This is the centripetal force.

Another friction force acts tangentially, increasing the tangential velocity. This force is coming out of the page of the back/front view Free Body Diagram.

answered Nov 25, 2018 by sammy gerbil (28,448 points)
selected Nov 30, 2018 by Jorge Daniel
NOTE: There are more new comments above.

Regarding this comment:

'Another friction force acts tangentially, increasing the tangential velocity. This force is coming out of the page of the back/front view Free Body Diagram.'

The fact that the tangential friction force increases the tangential velocity goes against my intuition. Actually, I thought about it as pointing against the tangential velocity (a friction force used to be thought as pointing against the movement direction)...
Friction forces oppose the tendency toward relative motion. The point of contact on the rear wheel would move backwards over the ground if there were no friction (the wheel turns without moving forward), so the friction force on the rear wheel is forwards. Without friction the point of contact onthe front wheel would move forwards because the cycle is being accelerated and the front wheel is not rotating fast enough - ie the 'no slip ' condition is broken because of the linear acceleration of the cycle. So the friction force on the front wheel is backwards.

The resultant friction force is forwards. It is the friction force which causes tangential acceleration. Without it the cycle would not accelerate.
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