Question

A particle of mass m is moving along a trajectory given by
\(x = x_0 + a\cos\omega_1 t\)
\(y = y_0 + b\sin\omega_2 t\)
The torque, acting on the particle about the origin, at \(t = 0\) is :

• A. Zero
• B. \(+my_0a\,\omega_1^2\,\hat{k}\)
• C. \(-m\left(x_0b\omega_2^2-y_0a\omega_1^2\right)\hat{k}\)
• D. \(m\left(-x_0b+y_0a\right)\omega_1^2\,\hat{k}\)

Hint:

Find the acceleration components at t = 0 and use \(\vec\tau=\vec r\times m\vec a\), i.e. \(\tau_z=m(xa_y-ya_x)\).

Answer 1

The Correct Option is B

\(a_x=\dfrac{d^2x}{dt^2}=-a\omega_1^2\cos\omega_1t\) and \(a_y=\dfrac{d^2y}{dt^2}=-b\omega_2^2\sin\omega_2t\).
At \(t=0\): \(a_x=-a\omega_1^2,\ a_y=0\); position \(x=x_0+a,\ y=y_0\).
\(\vec\tau=\vec r\times\vec F\Rightarrow \tau_z=m(x\,a_y-y\,a_x)=m\left[0-y_0(-a\omega_1^2)\right]=m\,y_0\,a\,\omega_1^2\).
So \(\vec\tau=+my_0a\omega_1^2\,\hat k\).
Answer: (2) \(+my_0a\,\omega_1^2\,\hat{k}\)