Figure shows a system of two masses m1 and m2 connected by a light inextensible string passing over a pulley having its rotational inertia I about its axis and radius R. The string does not slip over the pulley and the horizontal surface is smooth. Find the acceleration of the masses in m/s2. [Take : I = 3 kgm2, R = 1m, m1 = 1 kg, m2 = 1 kg ]

Engineering

Physics

Moment of Inertia

Constrained Relation

Newtons Second Law of Motion

Question

Figure shows a system of two masses m₁ and m₂ connected by a light inextensible string passing over a pulley having its rotational inertia I about its axis and radius R. The string does not slip over the pulley and the horizontal surface is smooth. Find the acceleration of the masses in m/s².
[Take : I = 3 kgm², R = 1m, m₁ = 1 kg, m₂ = 1 kg ]

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Since the string does not slip over the pulley, tangential acceleration at the rim a = αR, where a angular aceeleration, thus $α = \frac{a}{R}$ .
For the pulley,
torque $τ = (T_{1} - T_{2}) R = Iα = I \frac{a}{R}$

$\Rightarrow T_{1} - T_{2} = I \frac{a}{R^{2}}$

Equations of motion for m₁, m₂ and pulley are

m₁g – T₁ = m₁a

$T_{1} - T_{2} = I \frac{a}{R^{2}}$

T₂ = m₂a

Adding $m_{1} g = (m_{1} + m_{2} + \frac{1}{R^{2}}) a$

$\Rightarrow a = \frac{m_{1} g}{(m_{1} + m_{2} + \frac{1}{R^{2}})}$