If a cycle wheel of radius 4 , m completes on | vedclass

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(C) Given: Radius $r = 4 \, m$,Time period $T = 2 \, s$.Frequency $n = \frac{1}{T} = \frac{1}{2} \, Hz$.Since the wheel is rotating,the acceleration o

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$4{\pi ^2} \, m/s^2$

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(C) Given: Radius $r = 4 \, m$,Time period $T = 2 \, s$.Frequency $n = \frac{1}{T} = \frac{1}{2} \, Hz$.Since the wheel is rotating,the acceleration of a point on the rim is the centripetal acceleration $a_c$.The formula for centripetal acceleration is $a_c = \omega^2 r$,where $\omega = 2\pi n$.Substituting the values: $\omega = 2\pi 	imes \frac{1}{2} = \pi \, rad/s$.Therefore,$a_c = (\pi)^2 	imes 4 = 4\pi^2 \, m/s^2$.

If a cycle wheel of radius $4 \, m$ completes one revolution in two seconds,then the acceleration of a point on the rim of the cycle wheel will be:

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