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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

yashra singh , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 Years ago

m=0.36g=0.36*10-3
r=25cm=0.25m
angular acceleration = 1/3 rad/s2
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​2r.
=0.36*10-3*(2/3)2(0.25)
=0.04*10-3N
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