(A) The torque $\tau$ produced by the force $F$ acting at the rim of the wheel is given by $\tau = F \times R$. Given: Force $F = 10 \ N$,Radius $R =

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(A) The torque $\tau$ produced by the force $F$ acting at the rim of the wheel is given by $\tau = F \times R$. Given: Force $F = 10 \ N$,Radius $R = 0.2 \ m$,and angular acceleration $\alpha = 2 \ rad/s^2$. The relationship between torque,moment of inertia $I$,and angular acceleration $\alpha$ is $\tau = I \alpha$. Substituting the values: $F \times R = I \alpha$ $10 \times 0.2 = I \times 2$ $2 = 2I$ $I = 1 \ kg \ m^2$. Thus,the moment of inertia of the wheel is $1 \ kg \ m^2$.

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

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$A$ wheel of radius $0.2 \ m$ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $10 \ N$ as shown in the figure. The established torque produces an angular acceleration of $2 \ rad/s^2$. The moment of inertia of the wheel is . . . . . . $kg \ m^2$. (Acceleration due to gravity $= 10 \ m/s^2$)

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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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$A$ wheel rotates at $720 \ rpm$ about an axis passing through its center. It is brought to rest by applying a constant torque for $8 \ s$. What is the value of the torque in $Nm$? (Given: $I = \frac{24}{\pi} \ kg \cdot m^2$)

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