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Q.

A particle of mass m is moving along a trajectory given by x=x0+a cosω1t and y=y0+bsinω2t. The torque, acting on the particle about the origin, at t=0 is:

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a

+my0aω12k∧

b

m(−x0b+y0a)ω12k∧

c

−m(x0bw22−y0aw12)k^

d

Zero

answer is A.

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Detailed Solution

r→=(xo+acosω1t)i+(yo+bsinω2t)j; at t=0 r→=(xo+a)i+yoja→=d2xdt2i^+d2ydt2j^a→=(−aω12cosω1t)i^+(−bω22sinω2t)j; at t=0 a→=−aω12i^T=r→×F→=m(r→×a→)=m(yo(−aω12)−k^)=myoaω12k^
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A particle of mass m is moving along a trajectory given by x=x0+a cosω1t and y=y0+bsinω2t. The torque, acting on the particle about the origin, at t=0 is: