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Question

${\omega _1} = 20$ $rad/sec$,

${\omega _2} = 0,\,t = 4sec.$

So angular retardation $\alpha  = \frac{{{\omega _1} - {\omega _2}}}{t} = \frac

Vedclass · Accepted Answer

$30$

Vedclass · Answer

${\omega _1} = 20$ $rad/sec$,

${\omega _2} = 0,\,t = 4sec.$

So angular retardation $\alpha  = \frac{{{\omega _1} - {\omega _2}}}{t} = \frac{{20}}{4} = 5\ rad/se{c^2}$

Now angular speed after $2\ sec$,         

${\omega _2} = {\omega _1} - \alpha t$$ = 20 - 5 	imes 2$ = $10\,$$rad/sec$

Work done by torque in $2\ sec$ = loss in kinetic energy = $\frac{1}{2}I\,\left( {\omega _1^2 - \omega _2^2} ight)$$ = \frac{1}{2}(0.20)\,({(20)^2} - {(10)^2})$

$ = \frac{1}{2} 	imes \,\,0.2 	imes 300$= $30\ J.$

A wheel is rotating with an angular speed of $20\,rad/sec$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg-m^2$, then the work done by the torque in two seconds will be .......... $J$

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