Rotational Motion

JEE Main level, single correct, medium

17 of 280

Question 17 of 280

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 MM. Two blocks of mass of MM and mm 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).

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Answer: (B) (M−m)g[(83)M+m]\frac{(M-m) g}{\left[\left(\frac{8}{3}\right) M+m\right]}

The pulley consists of a rim (mass MM) and two rods (each mass MM) forming a diameter structure.
The moment of inertia is: I=Irim+Irods=MR2+2×MR23=5MR23I = I_{rim} + I_{rods} = MR^2 + 2 \times \frac{MR^2}{3} = \frac{5MR^2}{3}.
For the Atwood pulley system, applying Newton's second law:
Mass mm (upward): T1=m(g−a)T_1 = m(g-a)
Mass MM (downward): T2=M(g+a)T_2 = M(g+a)
Pulley rotation: T2R−T1R=Iα=5MR23⋅aRT_2 R - T_1 R = I\alpha = \frac{5MR^2}{3} \cdot \frac{a}{R}
Thus: T2−T1=5Ma3T_2 - T_1 = \frac{5Ma}{3}
Substituting: M(g+a)−m(g−a)=5Ma3M(g+a) - m(g-a) = \frac{5Ma}{3}
(M−m)g+(M+m)a=5Ma3(M-m)g + (M+m)a = \frac{5Ma}{3}
(M−m)g=a(5M3−M−m)=a(2M3−m)(M-m)g = a\left(\frac{5M}{3} - M - m\right) = a\left(\frac{2M}{3} - m\right)
a=(M−m)g2M−3m3=3(M−m)g2M+3ma = \frac{(M-m)g}{\frac{2M-3m}{3}} = \frac{3(M-m)g}{2M+3m}, which is option (2).

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