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

• hitemup
In summary, the conversation discusses the motion of a wheel with a given rotational inertia and initial angular speed as it is lowered to the ground. The wheel initially slips before beginning to roll without slipping. The length of time the wheel slips before rolling without slipping is determined by setting up equations for translational and rotational motion and using the condition that the final translational velocity is equal to the final angular velocity multiplied by the wheel's radius. The direction of friction is determined by the direction of the wheel's initial rotation.
hitemup

## 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 don't 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?

Last edited:

OldEngr63 said:
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.

Joshua L said:
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)

Last edited:
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.

Last edited:
hitemup

## 1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotational motion. It depends on the mass and the distribution of that mass around an axis of rotation.

## 2. How is rotational inertia calculated?

Rotational inertia is calculated using the formula I = MR2, where I is the moment of inertia, M is the mass of the object, and R is the distance from the axis of rotation to the mass.

## 3. What does it mean for a wheel to have a rotational inertia of 1/2MR2?

A wheel with a rotational inertia of 1/2MR2 means that the distribution of its mass is concentrated at the outer edge or rim of the wheel, with a radius of R. This results in the wheel having less resistance to changes in its rotational motion compared to a wheel with a different distribution of mass.

## 4. How does rotational inertia affect the performance of a wheel?

The rotational inertia of a wheel affects its acceleration and its ability to maintain its angular velocity. A wheel with a lower rotational inertia will accelerate and decelerate more quickly, while a wheel with a higher rotational inertia will maintain its angular velocity more easily.

## 5. Can rotational inertia be changed?

Yes, rotational inertia can be changed by altering the distribution of mass in an object or by changing the axis of rotation. For example, by moving the mass closer to the axis of rotation, the rotational inertia will decrease. Similarly, changing the axis of rotation can also affect the rotational inertia of an object.

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