First slide

Rigid body doing transational plus rotational motion

Question

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 $s^{- 1}$ . How much work has to be done to stop it?

Moderate

A

2 J
B

4 J
C

6 J
D

8 J

Solution

Here, R = 2m, M = 100 kg, v = 20 cm $s^{- 1}$
= ${20 × 10}^{- 2} {ms}^{- 1}$
Total kinetic of the loop = $K_{T} + K_{R}$
= $\frac{1}{2} {Mv}^{2} + \frac{1}{2} {Iω}^{2} [∴ For a hoop, I = {MR}^{2}]$
= $\frac{1}{2} {Mv}^{2} + \frac{1}{2} {MR}^{2} ω^{2}$
= $\frac{1}{2} {Mv}^{2} + \frac{1}{2} {Mv}^{2} [∵ v = Rω]$
= ${Mv}^{2}$
Work required to stop the hoop = Total kinetic energy of the hoop
= ${Mv}^{2} = (100 kg) {(20 \times 10^{- 2} {ms}^{- 1})}^{2} = 4 J$

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