Question 98: Centre of Mass and Rotational Motion

Question

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is ________ (assume no slipping of string on pulley).

Accepted Answer

For a hoop of mass $M$ and radius $R$, the moment of inertia is given as,
$$ \mathrm{I}_{\mathrm{rim}}=\mathrm{MR}^2 $$
Each rod has mass $M$ and length $L=2 R$. The moment of inertia of a rod rotating about its center is $\frac{1}{12} M L^2$.
$$ \mathrm{I}_{	ext {rod }}=2 	imes \frac{1}{12} \mathrm{M}(2 \mathrm{R})^2=\frac{8 \mathrm{MR}^2}{12}=\frac{2}{3} \mathrm{MR}^2 $$
Total moment of inertia of the system is,
$$ \mathrm{I}=\mathrm{I}_{	ext {rim }}+\mathrm{I}_{	ext {rod }}=\mathrm{MR}^2+\frac{2}{3} \mathrm{MR}^2=\frac{5}{3} \mathrm{MR}^2 $$
Let a be the linear acceleration of the blocks and $\alpha$ be the angular acceleration of the pulley. Assuming no slipping, $\mathrm{a}=\mathrm{R} \alpha$.
For Mass M,
$$ \mathrm{Mg}-\mathrm{T}_1=\mathrm{Ma} \ldots 	ext { (i) } $$
For Mass m,
$$ \mathrm{T}_2-\mathrm{mg}=\mathrm{ma} . $$
The net torque $	au$ about the centre of pulley is provided by the difference in tensions:
$$ 	au=\left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R}=\mathrm{I} \alpha $$
Substituting $\mathrm{I}=\frac{5}{3} \mathrm{MR}^2$ and $\alpha=\frac{\mathrm{a}}{\mathrm{R}}$ :
$$ \begin{aligned} \left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R} & =\left(\frac{5}{3} \mathrm{MR}^2ight) \frac{\mathrm{a}}{\mathrm{R}} \ \Rightarrow \mathrm{T}_1-\mathrm{T}_2 & =\frac{5}{3} \mathrm{Ma} \ldots(	ext { iii }) \end{aligned} $$
Adding equations (i) and (ii):
$$ \begin{aligned} & \left(M g-T_1ight)+\left(T_2-m gight)=M a+m a \\ & \Rightarrow (M-m) g-\left(T_1-T_2ight)=(M+m) a \end{aligned} $$
Substituting the value of ( $\mathrm{T}_1-\mathrm{T}_2$ ) from equation (iii) :
$$ (M-m) g-\frac{5}{3} M a=(M+m) a $$
$$\Rightarrow $$ $$(M-m) g=(M+m) a+\frac{5}{3} M a$$
$$\Rightarrow $$ $(M-m) g=\left(M+m+\frac{5}{3} Might) a$
$$\Rightarrow $$ $(M-m) g=\left(\frac{8}{3} M+might) a$
$$\Rightarrow $$ $$ a=\frac{(M-m) g}{\left(\frac{8}{3} M+might)} $$
Therefore, the acceleration of the blocks is $\frac{(M-m) g}{\left(\frac{8}{3} M+might)}$.
Hence, the correct option is (A).

Answer

$$
\dfrac{(M-m) g}{\left[\left(\frac{8}{3}\right) M+m\right]}
$$

Answer

$ \dfrac{(M - m)g}{2M + m} $

Answer

$ \dfrac{(M - m)g}{M + m} $

Answer

$$
\dfrac{(M-m) g}{\left[\left(\frac{13}{6}\right) M+m\right]}
$$

Centre of Mass and Rotational Motion

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