Answer to Question #92243 in Classical Mechanics for James Walker

Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes
an angle 60° with the horizontal. They start to roll without slipping at the same instant of
time along the shortest path. If the time difference between their reaching the ground is
(2 − √3) /√10

Answer 1

The acceleration of disc is given by formula

$$a=\frac{g}{\sqrt {3}} (1)$$

The acceleration of ring is given by formula

$$a=\frac{g×\sqrt {3}}{4} (2)$$

We can find time t

$$t=\sqrt { \frac{2S}{a}} (3)$$

where

$$S=\frac{h} {\sin 60°} (4)$$

The difference of time is equal to

$$Δt=\frac{2 - \sqrt {3}} {\sqrt {10}} (5)$$

We put (1) and (2) in (3) and we get using (5):

h=0.75 m

Answer

0.75 m