Find the velocity of a rolling disk at time t

[Krushnaraj Pandya]

Homework Statement

A solid circular disk (m,R) translates with v1 and rotates with angular velocity w1=2v1/R, at t=0. Because of friction, pure rolling starts at time t. the coefficient of friction is C. Find velocity of center of disk at time t

Homework Equations

1) for pure rolling v=Rw
2) for translational motion v=u+at
3) for rotational motion w=w1+(alpha)t where alpha is angular acceleration
4) kinetic friction= Cmg
5) Torque=Moment of inertia*alpha
6) Moment of Inertia of disk=(MR^2)/2

The Attempt at a Solution

using the above equations, V(f)=v1-Cgt and W(f)=V(f)/R={(2v1)/R}-(alpha)t, I found t to be v1/3Cg.
Again, using 2); v at time t=v1-Cgt, putting t I get the answer as 2v1/3 but my textbook says the answer is 4v1/3. Am I making a mistake somewhere?

 

[kuruman]

Am I making a mistake somewhere?

Yes, you are making the mistake that W(f)=V(f)/R (I assume this is $\omega=V/R$ ) when the wheel is rolling and slipping at the same time. This relation between angular speed and translational speed is valid only for wheels that roll without slipping. In fact the onset of rolling without slipping is when the angular speed becomes $V/R$.

 

[Nathanael]

@kuruman
I think “w(f)” means ω_f the final rolling rate.

@Krushnaraj Pandya
The problem is not clear. Is it initially spinning with 2V₁/R of back-spin or forward-spin? (Or maybe it’s a 2D problem with side spin!) Since you gave the answer, I can deduce it’s forward-spin, but that should be stated in some way.

As for your mistake... do you have a clear picture of the situation? Which way is friction directed?

 

[haruspex]

V(f)=v1-Cgt and W(f)=V(f)/R={(2v1)/R}-(alpha)t

Those two equations are not consistent with regard to this question:

back-spin or forward-spin?

 

[Delta2]

How do you calculate a and alpha??. I think your mistake is that you took both as negative, a should be positive but alpha negative.

 

[Krushnaraj Pandya]

@Nathanael w(f) indeed means final rolling rate, the problem had a simple figure (but I don't have a camera)- I have tried to include all the information, sorry if it is still somewhat unclear. The spin is in forward direction. The friction has to be acting backwards for rolling to start, since translational velocity has to decrease and rotational velocity has to increase. This means I made a mistake in posting one of the equations, the correct one is w(f)=V(f)/R={(2v1)/R}+(alpha)t, instead of the minus sign since rate of rolling is increasing. Despite that, I had used the correct equation in calculating time t, the expression for t is correct as per my textbook too. But the velocity of center of disk is given as 4v1/3, which does not agree with the answer when I write V of center of disk=v1-at. where a is Cg and t is the one I calculated (supposedly correctly) as v1/3Cg

 

[Krushnaraj Pandya]

How do you calculate a and alpha??. I think your mistake is that you took both as negative, a should be positive but alpha negative.

did you mean vice versa? since v is decreasing, shouldn't a be negative and alpha positive (as rolling rate should increase to stop sliding of the disk and start it rolling)

 

[Delta2]

The correct is V=V1+at. not -at.

 

[Krushnaraj Pandya]

The correct is V=V1+at. not -at.

Ah! yes, but since friction is backwards, it is equal to -Cg, I have written it directly with the minus sign so we only have to worry about the magnitude later.
Pardon if that caused any confusion

 

[Delta2]

did you mean vice versa? since v is decreasing, shouldn't a be negative and alpha positive (as rolling rate should increase to stop sliding of the disk and start it rolling)

If rolling rate is increased while translational velocity is decreased we will never reach the rolling without slipping condition.

 

[Nathanael]

@Krushnaraj Pandya
The ball is spinning forward at twice the rate of a rolling ball. If we have friction pointing backwards as you suggest, then the linear speed will decrease and the rotational speed will increase. Does this sound like it leads to rolling...?

 

[Krushnaraj Pandya]

@Krushnaraj Pandya
The ball is spinning forward at twice the rate of a rolling ball. If we have friction pointing backwards as you suggest, then the linear speed will decrease and the rotational speed will increase. Does this sound like it leads to rolling...?

OH!, I intuitively assumed that since velocity and angular velocity are in forward direction, the disk must be sliding forward

 

[Nathanael]

A perhaps better way to see that the friction is forward is that friction always opposes the relative motion. Since it’s spinning too fast forward, the bottom (contact point) of the ball tends to slide backwards, and friction opposes that by pointing forwards.

 

[Krushnaraj Pandya]

A perhaps better way to see that the friction is forward is that friction always opposes the relative motion. Since it’s spinning too fast forward, the bottom (contact point) of the ball tends to slide backwards, and friction opposes that pointing forwards.

Beautiful! I can visualize it easily now. Let me give the question a 'spin' again (pun intended) and see if I can get the answer.

 

[Krushnaraj Pandya]

How do you calculate a and alpha??. I think your mistake is that you took both as negative, a should be positive but alpha negative.

I got your point, I'll keep that in mind and try to solve it again

 

[Krushnaraj Pandya]

Got it! Also made a note of my mistake in my notebook so I don't make it again. Thank you everyone for helping, @Nathanael how is it you immediately know the mistake I must have made...grateful to have this forum.

 

[Nathanael]

@Nathanael how is it you immediately know the mistake I must have made...grateful to have this forum.

Haha, well, the better you explain your work, the easier it is to pinpoint mistakes!

This website really is great. Many regulars here have truly brilliant minds, and they regularly help people at all levels. I’m grateful as well. Thanks @Greg Bernhardt

 

[kuruman]

For future reference, you might consider conserving angular momentum about any point on the surface on which the wheel is rolling. Friction, gravity and normal force are the only external forces acting on the wheel and the net external torque is zero about any point on the surface. "Spin" angular momentum, $I_{cm}\omega$, about the center of mass is converted to "orbital" angular momentum about the surface, $mvR$, without loss. The details, like coefficient of friction, time to start rolling without sliding, etc. don't enter the picture as you can see from the final answer.

 

[Krushnaraj Pandya]

For future reference, you might consider conserving angular momentum about any point on the surface on which the wheel is rolling. Friction, gravity and normal force are the only external forces acting on the wheel and the net external torque is zero about any point on the surface. "Spin" angular momentum, $I_{cm}\omega$, about the center of mass is converted to "orbital" angular momentum about the surface, $mvR$, without loss. The details, like coefficient of friction, time to start rolling without sliding, etc. don't enter the picture as you can see from the final answer.

Ah! yes, that'd be fruitful...thank you for the insight