A ball is moving uniformly in a circular path | vedclass

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(D) The time period of the ball is $T = 1.5 \, s$. In $t = 8.3 \, s$,the number of revolutions completed is $n = \frac{8.3}{1.5} = 5.533$ revolutions.

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(D) The time period of the ball is $T = 1.5 \, s$. In $t = 8.3 \, s$,the number of revolutions completed is $n = \frac{8.3}{1.5} = 5.533$ revolutions. After $5$ complete revolutions $(t = 7.5 \, s)$,the ball returns to its initial position at $t = 0 \, s$. The remaining time is $\Delta t = 8.3 - 7.5 = 0.8 \, s$. The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \, rad/s$. The angle covered in $\Delta t = 0.8 \, s$ is $	heta = \omega \Delta t = \left(\frac{4\pi}{3}ight) 	imes 0.8 = \frac{3.2\pi}{3} \approx 1.067\pi \, rad$. This angle is slightly greater than $\pi \, rad$ $(180^\circ)$. Since the angle is approximately $192^\circ$,the ball is very close to the point diametrically opposite to the starting position. The displacement for a diameter is $2R = 2 	imes 1 = 2 \, m$. Thus,the magnitude of displacement is closest to $2 \, m$.

$A$ ball is moving uniformly in a circular path of radius $1 \, m$ with a time period of $1.5 \, s$. If the ball is suddenly stopped at $t = 8.3 \, s$,the magnitude of the displacement of the ball with respect to its position at $t = 0 \, s$ is closest to .......... $m$.

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