A small particle of mass 0.36 g rests on a horizontal turntable at a distance 25cm from the axis of spindle. The turntable is accelerated at a rate of Alpha = 1/3 Rad s^-2. The frictional force that the table exerts on the particle two seconds after the startup is

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Arun · Answer

m=0.36g=0.36*10 -3 r=25cm=0.25m angular acceleration = 1/3 rad/s 2 Angular velocity after 2 s is v. ω = 0+(1/3)*2 ω =2/3 rad/s So, we know that the cancelling force with friction is the centripetal force which acts on the body, which is equal to mω​ 2 r. =0.36*10 -3 *(2/3) 2 (0.25) =0.04*10 -3 N

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A small particle of mass 0.36 g rests on a horizontal turntable at a distance 25cm from the axis of spindle. The turntable is accelerated at a rate of Alpha = 1/3 Rad s^-2. The frictional force that the table exerts on the particle two seconds after the startup is

A small particle of mass 0.36 g rests on a horizontal turntable at a distance 25cm from the axis of spindle. The turntable is accelerated at a rate of Alpha = 1/3 Rad s^-2. The frictional force that the table exerts on the particle two seconds after the startup is

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