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# Rotational Work and Kinetic Energy

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A 0.050kg phonograph record with a radius of 0.15m drops onto a turntable and is soon rotating at 33 1/3 rpm. How much work must be supplied to get the record to rotate at this speed, and what supplies the work?

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A 0.050 kg phonograph record with a radius of 0.15m drops onto a turntable and is soon rotating at 33 1/3 rpm. How much work must be supplied to get the record to rotate at this speed, and what supplies the work?

KE of rotation= 1/2 (I w ^2)
where I= rotational inertia
w = angular speed

I of a disc= 1/2 (M r^2)
r= 0.15 m
M= 0.05 Kg
Therefore I = 1/2 M r^2= 0.0005625 kg ? m^2 =1/2 x 0.05x 0.15^2

w =2 pi f
f= 33 1/3 rpm
= 5/9 revolutions per second =100/3 /60

Therefore w = 3.4907 rad/s =2 pi x 5/9

And KE of rotation = 0.003427 Joules =1/2 x 0.0005625x 3.4907^2

Work supplied = Final Kinetic energy - Initial Kinetic energy= Final KE since Initial Kinetic energy=0
as the record starts from rest

Therefore work done= 0.003427 Joules
This work is supplied by the turntable

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