Verified Solution • 2025–26 Session moderate

QJEEM19Ph31d4: JEE Main 2019 Physics - Class 11 JEEM19Ph31d4 Solution: A particle of mass m is moving along a trajectory given by x =…

A particle of mass m is moving along a trajectory given by
x = x0 + a cos ω\omega 1t
y = y0 + b sin ω\omega 2t
The torque, acting on the particle about the origin, at t = 0 is :

Multiple Choice Options

  • A Zero
  • B +my0a ω12\omega _1^2 k^\widehat k
  • C −m(x0bω22−y0aω12)k^- m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k
  • D m (–x0b + y0a) ω12\omega _1^2 k^\widehat k

Step-by-Step Verified Solution

F→=ma→=m[−aω12cos⁡ω,ti^−bω22sin⁡ω2tj^]\overrightarrow F = m\overrightarrow a = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} \right]

f→t=0=−maω12i^{\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i

r→t=0=(X0+a)i^+yj^{\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j

τ→=r→×F→=my0aω12k^\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = m{y_0}a\omega _1^2\widehat k

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