(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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In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

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$A$ uniform disc rotates at $10$ revolutions per second. $A$ torque is applied to it, producing an angular acceleration of $5 \text{ rad s}^{-2}$. After $2 \text{ s}$, its angular velocity is ...... $\text{rad s}^{-1}$ and the number of revolutions completed by the disc in $2 \text{ s}$ is ...... .

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The moment of inertia of a wheel about its transverse axis is $2 \ kg \cdot m^2$ and it is rotating at $60 \ rpm$ about that axis. The torque required to stop the wheel in one minute is:

$A$ non-uniform rod $OA$ of linear mass density $\lambda = \lambda_0 x$ (where $\lambda_0$ is a constant) is suspended from the ceiling with a hinge joint at $O$ and a light string as shown in the figure. Find the angular acceleration of the rod just after the string is cut.

In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

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