A body of mass 1 ,kg is suspended by a weight | vedclass

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Question

(D) Given,mass of the body,$m_1 = 1 \,kg$.Mass of the pulley,$m_2 = 2 \,kg$.According to the law of conservation of energy,the potential energy lost b

Vedclass · Accepted Answer

$4 \,ms^{-1}$

Vedclass · Answer

(D) Given,mass of the body,$m_1 = 1 \,kg$.Mass of the pulley,$m_2 = 2 \,kg$.According to the law of conservation of energy,the potential energy lost by the mass is equal to the sum of the kinetic energy of the mass and the rotational kinetic energy of the pulley.$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{2} I \omega^2$Since the pulley is a disc,its moment of inertia is $I = \frac{1}{2} m_2 R^2$. Also,$\omega = \frac{v}{R}$.Substituting these values:$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{2} \left( \frac{1}{2} m_2 R^2 ight) \left( \frac{v}{R} ight)^2$$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{4} m_2 v^2$Substituting the given values $(m_1 = 1 \,kg, m_2 = 2 \,kg, g = 10 \,ms^{-2}, h = 1.6 \,m)$:$1 	imes 10 	imes 1.6 = \frac{1}{2} 	imes 1 	imes v^2 + \frac{1}{4} 	imes 2 	imes v^2$$16 = 0.5 v^2 + 0.5 v^2$$16 = v^2$$v = 4 \,ms^{-1}$

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$

$A$ body of mass $1 \,kg$ is suspended by a weightless string which passes over a pulley of mass $2 \,kg$ as shown in the figure. The mass is released from a height of $1.6 \,m$ from the ground. With what velocity does it strike the ground?

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