Question

An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attached to the other end of the rod. The pendulum thus formed is hung by the hoop onto a revolving shaft. The coefficient of friction between the shaft and the hoop is µ. Find the equilibrium angle between the rod and the vertical.

Answer 1

The equilibrium angle for the described pendulum setup would be where the gravitational torque caused by the ball's weight balances with the frictional torque from the hoop and revolving shaft.

Step-by-step explanation:

To determine the equilibrium angle between the rod and the vertical for a pendulum made of a wire rod, ball, and hoop on a revolving shaft with coefficient of friction µ, we need to consider the forces in play and the concept of torques. Unfortunately, without a specific diagram or more information, providing a precise answer is challenging. However, the equilibrium angle generally depends on the balance between the gravitational torque due to the ball's weight and the frictional torque due to the hoop's contact with the revolving shaft. The rod will settle at an angle where these two torques are equal and opposite, resulting in no net torque and a state of equilibrium.

Answer 2

`arcsin((R\mu)/((R+l)√(\mu^2+1)))`

Step-by-step explanation:

By the Law of Sines,

`sin \theta = (sin \phi R)/( l + R)`

From Newton's Law,

`mg = N√(\mu^2+1)`

And the last equation again from Newton's Law,

`\mu N = mgsin\phi`

Then if we collect all equations together,

`\mu N = mgsin\phi = N√(\mu^2+1)sin\phi\\`

`sin\theta = (\mu R)/( (l + R)√(\mu^2+1))`

Thus,

`\theta = arcsin((R\mu)/((R+l)√(\mu^2+1)))`