A pulley is hinged at the centre and a massle | vedclass

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(A) $1$. The constant force $F$ applied at the rim of the pulley creates a constant torque $	au = F \cdot R$ about the hinge,where $R$ is the radius

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its angular velocity increases,but force on hinge remains constant

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(A) $1$. The constant force $F$ applied at the rim of the pulley creates a constant torque $	au = F \cdot R$ about the hinge,where $R$ is the radius of the pulley.$2$. According to the rotational analogue of Newton's second law,$	au = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. Since $	au$ is constant,$\alpha$ is also constant.$3$. Starting from rest,the angular velocity $\omega$ at time $t$ is given by $\omega = \alpha t$. As $t$ increases,$\omega$ increases linearly.$4$. To maintain equilibrium in the translational direction,the hinge must exert a reaction force $F_h$ to balance the applied force $F$. Since the pulley is not moving translationally,the net force on the pulley must be zero. Therefore,the hinge force $F_h$ must be equal and opposite to the applied force $F$ at all times. Thus,the force on the hinge remains constant.

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force $F$ starting from rest. As the time increases,

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