Answer to Question #92243 in Classical Mechanics for James Walker

Question #92243
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
1
Expert's answer
2019-08-05T15:19:29-0400

The acceleration of disc is given by formula


a=g3(1)a=\frac{g}{\sqrt {3}} (1)

The acceleration of ring is given by formula


a=g×34(2)a=\frac{g×\sqrt {3}}{4} (2)

We can find time t

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

where

S=hsin60°(4)S=\frac{h} {\sin 60°} (4)

The difference of time is equal to


Δt=2310(5)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


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