- #1
palaphys
- 235
- 12
- Homework Statement
- A uniform disc of mass M and radius R is rotating freely about its central vertical axis with angular speed w0. Another disc of mass m and radius r is free to rotate about a horizontal rod AB. Length of the rod AB is 𝐿 ( < 𝑅 ) and its end A is rigidly attached to the vertical axis of the first disc. The disc of mass m, initially at rest, is placed gently on the disc of mass M as shown in figure. Find the time after which the slipping between the two discs will cease. Assume that normal reaction between the two discs is equal to mg. Coefficient of friction between the two discs is μ.
- Relevant Equations
- Tau= Ialpha , L=Iw
Consider the diagram shown above.
I came up with two ideas to solve this: i) I do not see any external torques about the vertical axis in this problem. so, angular momentum is conserved about this axis. (is it?)
ii)the standard way, that is, writing newton's second law for rotation, and applying the condition for slipping to cease, i.e the velocity of the contact point is zero.
let me show my working for the (ii)nd idea first:
assume after time t slipping ceases and pure rolling starts, and at that time angular velocity of horizontal disc is ω and the angular velocity of vertical disc is ω ′.
##
\begin{align}
(1)\quad & r\omega' = L\omega \\[6pt]
(2)\quad & \omega = \omega_0 - \frac{\mu m g L}{\tfrac{1}{2} M R^2}\, t \\[6pt]
(3)\quad & \omega' = \frac{L\omega}{r} = 0 + \frac{\mu m g r}{\tfrac{1}{2} m r^2}\, t \\[12pt]
& \text{From (2) and (3):} \\[6pt]
& \frac{L}{r}\left( \omega_0 - \frac{2 \mu m g L}{M R^2}\, t \right)
= 2\frac{\mu m g r^2}{m r^2}\, t \\[12pt]
\implies\quad
& t = \frac{M R^2 L \omega_0}{2 \mu g \left(M R^2 + m L^2\right)}
\end{align}
##
as expected, this yields the correct answer. however, the first method has some confusion.
first off, IF at all angular momentum is conserved, there will be a component of it which is NOT in the z direction(that is ## L_{about cm}##), for the smaller vertical disc.(using the fact that ##
L = L_{\text{cm}} + L_{\text{about cm}}##)
but if we ignore this and conserve angular momentum only about the z axis, I get the answer
##\begin{align}
t = \frac{M R^2 L \omega_0}{2 \mu g \left(M R^2 + 2m L^2\right)}
\end{align}##
which is incorrect for some reason.
Please help.
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