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A particle of mass m is moving along a trajectory given by $x = x_{0} + a \cos ω_{1} t$ and $y = y_{0} + b \sin ω_{2} t$ . The torque, acting on the particle about the origin, at t=0 is:

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+my0aω12k∧

m(−x0b+y0a)ω12k∧

−m(x0bw22−y0aw12)k^

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detailed solution

Correct option is A

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:

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Ready to Test Your Skills?

free study material

A particle of mass m is moving along a trajectory given by $x = x_{0} + a \cos ω_{1} t$ and $y = y_{0} + b \sin ω_{2} t$ . The torque, acting on the particle about the origin, at t=0 is: