Question

Rotation question. help please.

Answer 1

A disc of radius r is spun to an angular speed w about its axis and then imparted a horizontal velocity of magnitude wr/4(at t=0) with its plane remaining vertical. The coefficient of friction between the disc and plane is mu.The sense of rotation and direction of its linear speed is such that at its bottom most point they have same direction. The disc return to its initial point in time t=?

Answer 2

i've had a crack at this but it's really quite fiddly and i'd recommend a second opinion.....unless you already know the answer


the only force acting here, at least initially, is \( F = \mu mg\)

it will initially deccelerate the disc according to \(m \ddot x = - \mu mg\) and rotationally will also slow down the spinning according to \(I \ddot \theta = \frac{1}{2} m r^2 \ddot \theta = - \mu mg r\)

so \(\dot x = - \mu gt + \frac{\omega r}{4}, ~ ~ x = - \dfrac{\mu g}{2} t^2 + \frac{\omega r}{4} t\)


repeat for spinning to get \(\dot \theta = - \dfrac{2\mu g}{r} t + w\) and \( \theta = - \dfrac{\mu g}{r} t^2 + wt\)


set \(\dot x = 0\) and it stops moving at \(t = \frac{\omega r}{4\mu g}\)


set \(\dot \theta = 0\) and it should stop spinning at \(t = \frac{\omega r}{2\mu g}\)

so it is still skidding when it stops moving along the +ve x axis, and so it will begin to move in the negative direction under basically the same equations.

so if we just take a punt here and calculate \(x (\frac{\omega r}{2\mu g}) = - \dfrac{\mu g}{2} [\frac{\omega r}{2\mu g}]^2 + \frac{\omega r}{4} [\frac{\omega r}{2\mu g}] = 0\)


sadly that cannot be the answer, because there will be an intermediate period when the disc stopped skidding and just rolled at constant \(\dot x\) and \(\theta\)

that should happen, i think, when \(- r \dot \theta = \dot x\) or


\(- r [ - \dfrac{2\mu g}{r} t + w] = - \mu gt + \frac{\omega r}{4} \implies t = \dfrac{5 \omega r}{12 \mu g}\) ie just shy of the punt we took

check \( \dot x( \dfrac{5 \omega r}{12 \mu g} ) = - \mu g [\dfrac{5 \omega r}{12 \mu g} ] +\dfrac{\omega r}{4} = - \dfrac{\omega r}{6}\) and \(\dot \theta = \dfrac{\omega}{6}\)

at that point \(x (\dfrac{5 \omega r}{12 \mu g}) = \dfrac{5 \omega^2 r^2}{288 \mu g}\) and so it will roll the remaining distance in time t* where

\(t^* = \dfrac{\frac{5 \omega^2 r^2}{288 \mu g}}{\frac{\omega r}{6}} = \dfrac{5 \omega r}{48 \mu g}\)

giving total journey time of

\( \dfrac{5 \omega r}{12 \mu g} + \dfrac{5 \omega r}{48 \mu g} = \dfrac{25 \omega r}{48 \mu g} \)

hope that helps in some way....even if it's wrong, lol

Answer 3

If skidding does not stop, the answer is pretty simple.
Net force is -µmg, which is constant;
Initial velocity is vo = wo.R/4
So time to get to initial position is twice the time taken to change direction (v =0)

Hence \(\tau=\dfrac{2v_o}{\mu g}=\dfrac{\omega _oR}{2\mu g}\)

If skidding stops while in motion, it is a bit more complicated.
I'll check and get back to you later.

Answer 4

I also solved the problem and I find the very same result as IrishBoy's

Answer 5

Thanks to both of you. I got the ans.