Question

A particle is acted on by two torques about the origin: τ→1 has a magnitude of 8 N·m and is directed in the positive direction of the x axis, and τ→2 has a magnitude of 8.9 N·m and is directed in the negative direction of the y axis. In unit-vector notation, find d⁢ℓ→/d⁢t, where ℓ→ is the angular momentum of the particle about the origin.

Answer 1

To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.

For the given problem we have to

$T_i = 8Nm$

$T_j = -8.9Nm$

In this way the torque acting on the particle as a function of distance and time is,

$\tau = \frac{dL}{dt} = 8\hat{i}-8.9\hat{j}$

The net torque acting on the particle is

$\tau_{net} = \sqrt{T_i^2+T_j^2}$

$\tau_{net} = \sqrt{(8)^2+(-8.9)^2}$

$\tau_{net} = 11.967Nm$

PART B) The direction of the torque is given by,

$tan\theta = \frac{y}{x}$

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}(\frac{-8.9}{8})$

$\theta = -48.04\°$

Therefore the torque direction is 48.04° below the x axis.