A string is wrapped around the rim of a wheel | vedclass

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(D) Given: Moment of inertia $I = 0.20 \ kg \cdot m^2$,Radius $r = 20 \ cm = 0.2 \ m$,Force $F = 20 \ N$,Time $t = 5 \ s$,Initial angular velocity $\o

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Given: Moment of inertia $I = 0.20 \ kg \cdot m^2$,Radius $r = 20 \ cm = 0.2 \ m$,Force $F = 20 \ N$,Time $t = 5 \ s$,Initial angular velocity $\omega_0 = 0$.The torque $	au$ applied to the wheel is given by $	au = F 	imes r$.Substituting the values: $	au = 20 \ N 	imes 0.2 \ m = 4 \ N \cdot m$.We know that torque is also related to angular acceleration $\alpha$ by the equation $	au = I \alpha$.Therefore,$\alpha = \frac{	au}{I} = \frac{4 \ N \cdot m}{0.20 \ kg \cdot m^2} = 20 \ rad/s^2$.Using the equation of motion for rotation: $\omega = \omega_0 + \alpha t$.Substituting the values: $\omega = 0 + (20 \ rad/s^2) 	imes (5 \ s) = 100 \ rad/s$.

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