__Question 7.19__:

__Solution__:*r*= 2 m

Mass of the hoop,

*m*= 100 kg

Velocity of the hoop,

*v*= 20 cm/s = 0.2 m/s

Total energy of the hoop = Translational KE + Rotational KE

E

_{T}= (1/2)mv

^{2}+ (1/2) I ω

^{2}

Moment of inertia of the hoop about its centre,

*I*=

*mr*

^{2}

^{}E

_{T}= (1/2)mv

^{2}+ (1/2) (mr

^{2})ω

^{2}

^{}

But we have the relation, v = rω

∴ E

_{T}= (1/2)mv

^{2}+ (1/2)mr

^{2}ω

^{2}

= (1/2)mv

^{2}+ (1/2)mv

^{2 }= mv

^{2}

The work required to be done for stopping the hoop is equal to the total energy of the hoop.

∴ Required work to be done,

*W*=

*mv*

^{2 }= 100 × (0.2)

^{2 }= 4 J

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