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Question: A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass ha...

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?

A

2 J

B

4 J

C

6 J

D

8 J

Answer

4 J

Explanation

Solution

Here, R = 2 m, M = 100 kg, v = 20 cm

Total kinetic energy of the hoop, =

=12Mv2+12Iω2= \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 }

=12= \frac { 1 } { 2 } [For a hoop, I = MR2\mathrm { MR } ^ { 2 } ] =12Mv2+12Mv2= \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } []

Work required to stop the hoop = total kinetic energy of the hoop .

=Mv2=(1000 kg)(20×10−2 m s−1)2=4 J= \mathrm { Mv } ^ { 2 } = ( 1000 \mathrm {~kg} ) \left( 20 \times 10 ^ { - 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 } \right) ^ { 2 } = 4 \mathrm {~J}