Systems of Particles and Rotational Motion
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Question : 19
Total: 33
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm s– 1
Solution:
Radius of hoop, R = 2 m
Mass of hoop,M = 100 k g
Velocity of centre of mass= 20 c m s – 1 = 0.2 m s – 1
The total kinetic energy of the hoop
=
M v 2 +
I ω 2
=
M v 2 +
M R 2 ω 2
=
M v 2 +
M R 2 ω 2
=
M v 2 +
M v 2
= M v 2 = 100 × ( 0.2 ) 2 = 4 J
By work-energy theorem,
Work required to stop the hoop = Total kinetic energy of hoop = 4 J
Mass of hoop,
Velocity of centre of mass
The total kinetic energy of the hoop
By work-energy theorem,
Work required to stop the hoop = Total kinetic energy of hoop = 4 J
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