(D) Initial angular velocity $\omega_0 = 1200\text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi\text{ rad/s}$.Angular acceleration $\alpha = \frac{\Del

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(D) Initial angular velocity $\omega_0 = 1200\text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi\text{ rad/s}$.Angular acceleration $\alpha = \frac{\Delta\omega}{\Delta t} = \frac{0 - 40\pi}{10} = -4\pi\text{ rad/s}^2$.Moment of inertia of a solid sphere $I = \frac{2}{5}mR^2 = \frac{2}{5} \times 5 \times (0.04\text{ m})^2 = 2 \times 0.0016 = 0.0032\text{ kg m}^2$.Torque applied $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi\text{ Nm}$.Angular displacement $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 40\pi(10) + \frac{1}{2}(-4\pi)(10)^2 = 400\pi - 200\pi = 200\pi\text{ rad}$.Number of rotations $N = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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$A$ solid sphere of radius $4\text{ cm}$ and mass $5\text{ kg}$ is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of $1200\text{ rpm}$. It is brought to rest in $10\text{ s}$ 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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A
$0.128\pi\text{ Nm}$, $100$
B
$0.0128\pi\text{ Nm}$, $50$
C
$0.128\pi\text{ Nm}$, $50$
D
$0.0128\pi\text{ Nm}$, $100$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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