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Question

A small rubber wheel is used to drive a large pottery wheel, and they are mounted so that their circular edges touch. Assume that the small wheel has a radius of 2.0 cm and accelerates at the rate of 7.2 rad/s2, and it is in contact with the pottery wheel (radius 25.0 cm) without slipping.

1)Calculate the angular acceleration of the pottery wheel. (rad/s2)
2)Calculate the time it takes the pottery wheel to reach its required speed of 61 rpm.(s)

Answer 1

Since the wheels are in non-slipping contact, every revolution of the small wheel, or 2*pi radians, the larger wheel rotates
(2/25)*2*pi radians

so, lets look at theta for the small wheel
ths=.5*α*t^2
and th of the large
thl=(2/25)*.5*α*t^2
since w=dth/dt
wl=(2/25)*α*t
and
αl=dwl/dt
αl=(2/25)*α
1) (2/25)*7.2
0.58 rad/s^2

2) 1 rpm=60/(2*pi) rad/sec
61*30/3.14=.58*t
t=10 seconds

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