# A wheel with rotational inertia I = 1/2MR^2

## Homework Statement

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A wheel with rotational inertia I = 1/2MR^2 about its horizontal central axle is set spinning with initial angular speed W0. It is then lowered, and at the instant its edge touches the ground the speed of the axle (and CM) is zero. Initially the wheel slips when it touches the ground, but then begins to move forward and eventually rolls without slipping.
How long does the wheel slip before it begins to roll without slipping?

V = v0+ at
f = ma
T = I*alpha
W = W0 + alpha*t
W*r = V

## The Attempt at a Solution

Translational motion
Fs = ma
-kmg = ma
a = -kg
V = v0 + at
V = -kgt

Rotational motion
Fs*r = I*alpha
kmgr = 1/2mr^2*alpha
alpha = 2kg/r
W = W0 + alpha*t

Rolling without slipping condition

W*r = V

W0*r + 2kgt = -kgt
-3kgt = W0r
t = -W0r/(3kg)

My answer is correct if you dont see the minus sign. That probably means I couldn't draw the free body diagram properly. What should be the motion and the friction direction of the ball while it is slipping?

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OldEngr63
Gold Member

Sorry about that, it is coefficient of friction. (Mu)

This is a very interesting question. When the wheel first touches the ground, it's translational velocity is zero. The wheel also has a given angular velocity; let's say that it is clockwise. One can view that the wheel is slipping on the ground to the left, i.e. the wheel's edge that is sliding on the ground is moving leftwards. Therefore, there is a frictional force acting in the rightwards direction. During the event that we are focusing on, the wheel slips with and ends up having a translational velocity in a certain direction. There must have some force that changes the wheel's translational momentum in that particular way. The only significant forces that act on the wheel are the normal force, the gravitational force, and friction. Since a free-body diagram can show that the normal force and the gravitational force can cancel each other out, the frictional force must be the force that changes the wheel's translational momentum in such a way described above. The frictional force is in the same direction as the final velocity if the initial velocity is zero (To the right).

I'm not going to give you the answer to the problem directly, but hopefully this description helps answer your particular question.

This is a very interesting question. When the wheel first touches the ground, it's translational velocity is zero. The wheel also has a given angular velocity; let's say that it is clockwise. One can view that the wheel is slipping on the ground to the left, i.e. the wheel's edge that is sliding on the ground is moving leftwards. Therefore, there is a frictional force acting in the rightwards direction. During the event that we are focusing on, the wheel slips with and ends up having a translational velocity in a certain direction. There must have some force that changes the wheel's translational momentum in that particular way. The only significant forces that act on the wheel are the normal force, the gravitational force, and friction. Since a free-body diagram can show that the normal force and the gravitational force can cancel each other out, the frictional force must be the force that changes the wheel's translational momentum in such a way described above. The frictional force is in the same direction as the final velocity if the initial velocity is zero (To the right).

I'm not going to give you the answer to the problem directly, but hopefully this description helps answer your particular question.

Would the friction be leftwards if it was rotating counter clock wise while slipping? The question also asks the direction of the friction, and the correct answer of that is exactly what you have done. Does it mean that this wheel is absolutely rotating clockwise, so we cannot assume its direction counter clock wise?

I also managed to get the correct result thanks to you.

friction ---> rightwards
rotation ---> rightwards

Translational motion:

Mu*m*g = m*a
a = Mu*g (positive since the wheel is also moving right)
V_final = Mu*g*t

Rotational Motion

-Mu*m*g*r = 1/2*m*r^2*alpha (torque)
alpha = -2*Mu*g/r (negative since they have different directions of rotations.)
W_final = W_initial - 2*Mu*g*t/r

Condition

V_final = W_final *r
W_initial*r - 2*Mu*g*t = Mu*g*t
3*Mu*g*t = W_initial*r
t = W_initial*r/(3*Mu*g)

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• Joshua L
The frictional force is opposing the wheel's edge sliding leftwards on the ground. That's what friction tends to do; it opposes the inertial movement of a particle or body.

Yes, friction would be leftwards. It also doesn't matter what direction the initial angular velocity is. If it was initially rotating counter-clockwise, then the wheel's edge would be sliding on the ground rightwards and its corresponding frictional force would be acting leftwards. Thus, its final translational velocity would be to the left. Its initial direction wouldn't matter; the time of sliding would still be the same. So, it's up to the solver the choose an initial direction. This doesn't have to be done, but it would be easier in my opinion.

You have approached this problem with scalars, which is alright. You did solve the problem. The important piece of information that is used to solve this problem (in my opinion) is the realization that the frictional force is in the same direction as the wheel's translational acceleration, yet it opposes the spinning of the wheel.

I would have personally used vectors, but that is just a preference. I'm glad I could help.

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• hitemup