(A) Given: Diameter $D = 0.8 \ m$,so radius $R = 0.4 \ m$. Mass $M = 4 \ kg$. Torque $\tau = 2.56 \ N \ m$.For a circular disc,the moment of inertia a

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(A) Given: Diameter $D = 0.8 \ m$,so radius $R = 0.4 \ m$. Mass $M = 4 \ kg$. Torque $\tau = 2.56 \ N \ m$.For a circular disc,the moment of inertia about its central axis is $I = \frac{1}{2} M R^2$.$I = \frac{1}{2} \times 4 \times (0.4)^2 = 2 \times 0.16 = 0.32 \ kg \ m^2$.Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:$\alpha = \frac{\tau}{I} = \frac{2.56}{0.32}$.$\alpha = 8 \ rad \ s^{-2}$.

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

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$A$ circular disc of diameter $0.8 \ m$ and mass $4 \ kg$ is rolling on a smooth horizontal plane. If $2.56 \ N \ m$ torque is acting on the disc,then its angular acceleration is

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A
$8 \ rad \ s^{-2}$
B
$4 \ rad \ s^{-2}$
C
$2 \ rad \ s^{-2}$
D
$16 \ rad \ s^{-2}$

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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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$A$ circular disc of diameter $0.8 \ m$ and mass $4 \ kg$ is rolling on a smooth horizontal plane. If $2.56 \ N \ m$ torque is acting on the disc,then its angular acceleration is

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