Question

A particle is acted on by 2 torques about the origin:

has a magnitude of 2.0Nm and is directed inthe positive direction
of the x axis, and
has a magnitude of 4.0Nm and is directed inthe negative direction
of the y axis. In unit-vectornotation, find
where
is the angular momentum of the particleabout the origin.

Answer 1

We will start with the understanding that the first derivative of the angular momentum is equivalent to the torque, therefore,

`\frac{d\vec{l}}{dt} = \vec{\tau_(net)}`

Vectorially the given value can be expressed as,

`\tau_(net) = \vec{\tau_x}+\vec{\tau_y}`

`\tau_(net) = 2.0\hat{i}+4.0\hat{-j}`

This will allow us to find the magnitude of the total torque by the trigonometric properties of the vectors, for which:

`|\tau_(net)| = √(2^2+16^2)`

`|\tau_(net)| = 5√(2)N\cdot m`

Finally the angle will be given by the relation of the tangent,

`tan\theta = (4)/(2)`

`\theta = 63.4\°`

Direction of 63.4° clockwise from x-direction