Rotational Motion

JEE Main level, single correct, medium

6 of 280

Question 6 of 280

A solid sphere of radius 44 cm and mass 55 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 12001200 rpm. It is brought to rest in 1010 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _______ and _______ respectively.

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Answer: (D) 0.0128π0.0128\pi Nm, 100100

Initial angular velocity ω0=1200 rpm=1200×2π60=40π rad/s\omega_0 = 1200 \text{ rpm} = \dfrac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}

Final angular velocity ω=0\omega = 0

Time t=10 st = 10 \text{ s}

Angular acceleration α=ω−ω0t=0−40π10=−4π rad/s2\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{0 - 40\pi}{10} = -4\pi \text{ rad/s}^2

Moment of inertia of the solid sphere I=25MR2=25×5×(0.04)2=0.0032 kg m2I = \dfrac{2}{5}MR^2 = \dfrac{2}{5} \times 5 \times (0.04)^2 = 0.0032 \text{ kg m}^2

Torque τ=I∣α∣=0.0032×4π=0.0128π Nm\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi \text{ Nm}

Angle rotated θ=ω0+ω2t=40π+02×10=200π rad\theta = \dfrac{\omega_0 + \omega}{2}t = \dfrac{40\pi + 0}{2} \times 10 = 200\pi \text{ rad}

Number of rotations N=θ2π=200π2π=100N = \dfrac{\theta}{2\pi} = \dfrac{200\pi}{2\pi} = 100

Answer: 0.0128π0.0128\pi Nm, 100100

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