JEE MAIN - Physics (2011 - No. 22)
A pulley of radius $$2$$ $$m$$ is rotated about its axis by a force $$F = \left( {20t - 5{t^2}} \right)$$ newton (where $$t$$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $$10kg$$-$${m^2}$$ the number of rotation made by the pulley before its direction of motion is reversed, is:
more than $$3$$ but less than $$6$$
more than $$6$$ but less than $$9$$
more than $$9$$
less than $$3$$
Explanation
Given $$F = 20t - 5{t^2}$$, R = 2 m and $$I$$ = 10 kg m2
Torque applied on pulley $$\tau = FR$$
$$\therefore$$ $$\alpha = {{FR} \over I}$$ [ as $$\tau = I\alpha $$ ]
$$ \Rightarrow $$ $$\alpha = {{\left( {20t - 5{t^2}} \right) \times 2} \over {10}}$$
$$ \Rightarrow $$ $$\alpha = 4t - {t^2}$$
$$ \Rightarrow {{d\omega } \over {dt}} = 4t - {t^2}$$
$$ \Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)} dt$$
$$ \Rightarrow \omega = 2{t^2} - {{{r^3}} \over 3}$$
( At $$t = 0,6 \,s$$ $$\omega = 0$$ )
$$\omega = {{d\theta } \over {dt}} = 2{t^2} - {{{t^3}} \over 3}$$
$$\int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - {{{r^3}} \over 3}} \right)} dt$$
$$ \Rightarrow \theta = 36rad\,\, \Rightarrow n = {{36} \over {2\pi }} < 6$$
Torque applied on pulley $$\tau = FR$$
$$\therefore$$ $$\alpha = {{FR} \over I}$$ [ as $$\tau = I\alpha $$ ]
$$ \Rightarrow $$ $$\alpha = {{\left( {20t - 5{t^2}} \right) \times 2} \over {10}}$$
$$ \Rightarrow $$ $$\alpha = 4t - {t^2}$$
$$ \Rightarrow {{d\omega } \over {dt}} = 4t - {t^2}$$
$$ \Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)} dt$$
$$ \Rightarrow \omega = 2{t^2} - {{{r^3}} \over 3}$$
( At $$t = 0,6 \,s$$ $$\omega = 0$$ )
$$\omega = {{d\theta } \over {dt}} = 2{t^2} - {{{t^3}} \over 3}$$
$$\int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - {{{r^3}} \over 3}} \right)} dt$$
$$ \Rightarrow \theta = 36rad\,\, \Rightarrow n = {{36} \over {2\pi }} < 6$$
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