The length of the second's hand in a watch is | vedclass

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(D) The length of the second's hand is $r = 1 \, cm$. The time period of the second's hand is $T = 60 \, s$.The angular velocity is $\omega = \frac{2\

Vedclass · Accepted Answer

$\frac{\pi\sqrt{2}}{30} \, cm/sec$

Vedclass · Answer

(D) The length of the second's hand is $r = 1 \, cm$. The time period of the second's hand is $T = 60 \, s$.The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = \frac{\pi}{30} \, rad/s$.The linear velocity of the tip is $v = r\omega = 1 	imes \frac{\pi}{30} = \frac{\pi}{30} \, cm/s$.In $15 \, seconds$, the second's hand rotates through an angle $	heta = 90^\circ$ (since $60 \, s$ corresponds to $360^\circ$).The change in velocity is given by $\Delta v = |\vec{v_2} - \vec{v_1}| = 2v \sin(	heta/2)$.Substituting the values: $\Delta v = 2 	imes \left(\frac{\pi}{30}ight) 	imes \sin(90^\circ/2) = 2 	imes \frac{\pi}{30} 	imes \sin(45^\circ) = 2 	imes \frac{\pi}{30} 	imes \frac{1}{\sqrt{2}} = \frac{\pi\sqrt{2}}{30} \, cm/s$.

The length of the second's hand in a watch is $1 \, cm$. The change in velocity of its tip in $15 \, seconds$ is

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