The magnitude of torque on a particle of mass $1\,kg$ is $2.5\, Nm$ about the origin. If the force acting on it is $1\, N$, and the distance of the particle from the origin is $5\,m$, the angle between the force and the position vector is (in radians) :

Question

A uniform circular disc of mass 2 kg and radius 0.5 m is mounted on a frictionless axle. A force of 4 N is applied tangentially at the rim for 2 seconds. Find the angular velocity acquired by the disc at the end of 2 seconds.

Answer 1

To find the angle between the force vector and the position vector, we need to use the formula for torque in vector form. The torque $ \tau $ is given by:

\[\tau = \vec{r} \times \vec{F}\]

Here, $ \vec{r} $ is the position vector, $ \vec{F} $ is the force vector, and $ \times $ represents the vector cross product. The magnitude of the torque can be written as:

\[\tau = r \cdot F \cdot \sin\theta\]

Where:

$ r = 5 \, \text{m} $ (distance from origin)
$ F = 1 \, \text{N} $ (force magnitude)
$ \theta $ is the angle between the force and the position vector
$ \tau = 2.5 \, \text{Nm} $ (given torque magnitude)

Substituting the known values, we have:

\[2.5 = 5 \cdot 1 \cdot \sin\theta\]

\[2.5 = 5 \cdot \sin\theta\]

Solving for $ \sin\theta $, we get:

\[\sin\theta = \frac{2.5}{5} = 0.5\]

The angle $ \theta $ whose sine is 0.5 is:

\[\theta = \frac{\pi}{6}\] radians

Thus, the angle between the force and the position vector is indeed \[\frac{\pi }{6}\] radians.

Answer 2