(A) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad \,s^{-2}$.The moment of inertia of a disc about its central

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(A) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad \,s^{-2}$.The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.$I = \frac{1}{2} \times 1 \,kg \times (0.4 \,m)^2 = 0.5 \times 0.16 = 0.08 \,kg \,m^2$.The torque $\tau$ is given by $\tau = I \alpha$.$\tau = 0.08 \,kg \,m^2 \times 10 \,rad \,s^{-2} = 0.8 \,N \,m$.The tangential force $F$ applied at the rim is related to torque by $\tau = F \times R$.Therefore,$F = \frac{\tau}{R} = \frac{0.8 \,N \,m}{0.4 \,m} = 2 \,N$.

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

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$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is $10 \,rad \,s^{-2}$. The tangential force applied to the rim of the disc is (in $\,N$)

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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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$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is $10 \,rad \,s^{-2}$. The tangential force applied to the rim of the disc is (in $\,N$)

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