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Question

(B) Since the angular acceleration $\alpha$ is constant and the initial angular velocity $\omega_0 = 0$,the angle $	heta$ rotated in time $t$ is give

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Since the angular acceleration $\alpha$ is constant and the initial angular velocity $\omega_0 = 0$,the angle $	heta$ rotated in time $t$ is given by $	heta = \frac{1}{2} \alpha t^2$.Given $	heta = 15^{\circ}$ for time $t$,we have $15^{\circ} = \frac{1}{2} \alpha t^2$ (Equation $1$).Now,we need to find the angle rotated in the next $2t \ s$. The total time elapsed is $t + 2t = 3t \ s$.The total angle $	heta_{total}$ rotated in time $3t$ is $	heta_{total} = \frac{1}{2} \alpha (3t)^2 = 9 	imes (\frac{1}{2} \alpha t^2)$.Substituting Equation $1$ into this,we get $	heta_{total} = 9 	imes 15^{\circ} = 135^{\circ}$.The increase in angle in the next $2t \ s$ is $\Delta 	heta = 	heta_{total} - 	heta = 135^{\circ} - 15^{\circ} = 120^{\circ}$.

$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

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