(A) Given: Moment of inertia $I = 0.40 \ kg \cdot m^2$,radius $R = 10 \ cm = 0.1 \ m$,force $F = 40 \ N$,time $t = 10 \ s$,initial angular velocity $\

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(A) Given: Moment of inertia $I = 0.40 \ kg \cdot m^2$,radius $R = 10 \ cm = 0.1 \ m$,force $F = 40 \ N$,time $t = 10 \ s$,initial angular velocity $\omega_0 = 0 \ rad/s$.The torque $\tau$ applied to the wheel is given by $\tau = F \times R$.$\tau = 40 \times 0.1 = 4 \ N \cdot m$.Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:$4 = 0.40 \times \alpha \Rightarrow \alpha = \frac{4}{0.40} = 10 \ rad/s^2$.Using the equation of motion $\omega = \omega_0 + \alpha t$:$\omega = 0 + (10 \times 10) = 100 \ rad/s$.Thus,$x = 100$.

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

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$A$ string is wrapped around the rim of a wheel of moment of inertia $0.40 \ kg \cdot m^2$ and radius $10 \ cm$. The wheel is free to rotate about its axis. Initially,the wheel is at rest. The string is now pulled by a force of $40 \ N$. The angular velocity of the wheel after $10 \ s$ is $x \ rad/s$,where $x$ is $\qquad$

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$100$
B
$199$
C
$198$
D
$99$

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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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$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

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