A particle of mass | RIGID BODY DYNAMICS

JEE MAIN _2019_

S1_100419

Question

A particle of mass $m$ is moving along a trajectory given by $$ \begin{aligned} & x=x_0+a \cos \omega_1 t \\ & y=y_0+b \sin \omega_2 t \end{aligned} $$ The torque, acting on the particle about the origin, at $t=0$ is.

Select the correct option:

$-m\left(x_0 b \omega_2^2-y_0 a \omega_1^2\right) \hat{k}$

$m\left(-x_0 b+y_0 a\right) a_1^2 \hat{k}$

$+m y_0 a \omega_1^2 \hat{k}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Medium

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