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. How much work has to be done to stop it?
Answers
Answered by
140
Given ,
Mass of hoop = 100 Kg
Radius = 2m
Speed of centre of mass = 20cm/s = 0.2 m/s
Work done by hoop to stop it = total kinetic energy of the hoop
= (1/2) mv² + (1/2)Iw²
Moment of inertia of hoop = mr² and w = v/r
Then,
Work done =(1/2)mv² + (1/2)mr²×v²/r² = mv²
= 100 × (0.20)²
= 100 × 0.04 = 4 joule
Mass of hoop = 100 Kg
Radius = 2m
Speed of centre of mass = 20cm/s = 0.2 m/s
Work done by hoop to stop it = total kinetic energy of the hoop
= (1/2) mv² + (1/2)Iw²
Moment of inertia of hoop = mr² and w = v/r
Then,
Work done =(1/2)mv² + (1/2)mr²×v²/r² = mv²
= 100 × (0.20)²
= 100 × 0.04 = 4 joule
Answered by
39
Here, R = 2 m, M = 100 kg
v = 20 cm/s = 0.2 m/s
Total energy of the hoop =1/2Mv2 + 1/2Iw2
=1/2Mv2 + 1/2(MR2)w2
=1/2Mv2 +1/2Mv2 =Mv2
Work required to stop the hoop = total energy of the hoop W = Mv2 = 100 (0.2)2= 4 Joule.
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