A uniform disc is acted by 2 equal forces of magnitude f.one of them,acts tangentially to the disc, while other one is acting at the central point of the disc. the friction between disc surface and ground surface is nf. if r be the radius of the disc,then the calue of n would be
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Solution:-
let :-
》☆ fr be the friction exerting between disc surface and ground surface , then the motion of disc.
we have ,
》 2F -fr = ma...............(1)
》 and ( F + fr)r = Iw ( here w = omega)
》>>>> (F+fr)r = (1/2)mr^2×a/r........(2)
》here, a is the linear acceleration of the disc.
》from eqution (1) and (2)
》 (fr = 0) ans
☆i hope it help☆
let :-
》☆ fr be the friction exerting between disc surface and ground surface , then the motion of disc.
we have ,
》 2F -fr = ma...............(1)
》 and ( F + fr)r = Iw ( here w = omega)
》>>>> (F+fr)r = (1/2)mr^2×a/r........(2)
》here, a is the linear acceleration of the disc.
》from eqution (1) and (2)
》 (fr = 0) ans
☆i hope it help☆
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