If a cycle wheel of radius 4 m completes one revolution in two seconds, then acceleration of a point on the cycle wheel will be-(1)π^2m/s^2(2)2π^2m/s^2(3)4π^2m/s^2(4)8π^2m/s^2

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Vikas TU · Answer

angular velocity would be: w = theta/time = 2*pi/2 = pi rad accln. = w^R = > pi^2 * 4 = 4*pi^2 m/s^2.

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If a cycle wheel of radius 4 m completes one revolution in two seconds, then acceleration of a point on the cycle wheel will be-(1)π^2m/s^2(2)2π^2m/s^2(3)4π^2m/s^2(4)8π^2m/s^2

If a cycle wheel of radius 4 m completes one revolution in two seconds, then acceleration of a point on the cycle wheel will be-(1)π^2m/s^2(2)2π^2m/s^2(3)4π^2m/s^2(4)8π^2m/s^2

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If a cycle wheel of radius 4 m completes one revolution in two seconds, then acceleration of a point on the cycle wheel will be-(1)π^2m/s^2(2)2π^2m/s^2(3)4π^2m/s^2(4)8π^2m/s^2

If a cycle wheel of radius 4 m completes one revolution in two seconds, then acceleration of a point on the cycle wheel will be-(1)π^2m/s^2(2)2π^2m/s^2(3)4π^2m/s^2(4)8π^2m/s^2

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