a ring of diameter 1 meter is rotating with angular momentum of 10 J-s For increasing its angular momentum by 50% in one second required tangential force will be
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L = 10 J s. dL = 10/2 J-s
Assume that the ring is rotating around the axis through its center.
Torque = F * R = dL/ dt
F * 1/2 m = 5 J-s / 1 s
F = 10 N
Assume that the ring is rotating around the axis through its center.
Torque = F * R = dL/ dt
F * 1/2 m = 5 J-s / 1 s
F = 10 N
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