# Angular rotation of a wheel that slips

1. Dec 1, 2013

### g-racer

A wheel spinning clockwise on its axis at with angular velocity ω0 drops to the horizontal ground. It initially has no center-of-mass velocity. The coeﬃcient of kinetic friction between the ground and the barrel is µ. The radius of the wheel is R, and it is a solid disc of mass M. Express your answer in terms of the variables ω0, µ, etc. In part (e), you will be given some test values with which to calculate numbers using those answers.

(a) Calculate the friction force by the ground on the wheel and the torque (about the center of the wheel) by this force on the wheel. Draw a diagram of the wheel at the moment it hits the ground, showing the direction of the motion of the bottom of the wheel and the direction of the friction force.

(b) Calculate the moment of inertia of the wheel and use the linear and angular second laws to calculate the linear and angular accelerations of the wheel once the wheel is on the ground, but before it rolls smoothly. NOTE that they are not related by a = rα when the wheel is not rolling without slipping. Also be careful about the signs. If the initial rotation is taken as positive, what is the sign of the angular acceleration?
(c) Find the time and distance traveled from when the wheel ﬁrst hits the ground to when it ﬁrst rolls smoothly without slipping. For this, you will need the equations of motion, and to be careful about the signs of the accelerations. From this ﬁnd how fast the wheel is moving (center of mass linear velocity) when it does roll smoothly

For part a) F=ukmg and T=ukmgR
b) I=1/2MR^2. I think the angular acceleration will be T/I = 2ukg/R
is the angular accleration ukg I am not sure how to answer the rest of the question as we have only dealt with rolling with friction.

2. Dec 1, 2013

### voko

Your answer to b) indicates that the angular acceleration is positive. That means the disk will spin faster as it hits the ground. Does that sound reasonable?

For linear acceleration, use the force of friction you found earlier.

3. Dec 1, 2013

### g-racer

ok so angular acceleration = -2ukg/R and linear acceleration = ukmg/m=ukg ? I am not sure where to go from here for part c

4. Dec 2, 2013

### voko

You need to formulate the condition for "no slipping". What does "rolling without slipping" mean mathematically?

5. Dec 2, 2013

### g-racer

when rolling without slipping v=wr and d=theta.r but I am not sure how to work with this. thanks

6. Dec 2, 2013

### g-racer

I think I might have got it. so wf=wi+αt=wi-2ukgt/R and vf=vi+at=0+ukgt and as vf equals wf.r when they are not slipping we can set the equations equal for wf and solve for t which gives t=wiR/3ukg??? thanks

7. Dec 3, 2013

### voko

That looks good, well done!