Question

A-circular loop of Radius R is rotating with angular velocity w around its centre. A mass of m is hung from rim with length R of string making angle $\theta$ with vertical what is the value of w.

Answer 1

$\omega = \sqrt{\frac{g \tan\theta}{R(1 + \sin\theta)}}$

Answer 2

The problem describes a circular loop of radius R rotating with angular velocity $\omega$ around its center. A mass $m$ is hung from the rim of this loop by a string of length R. The string makes an angle $\theta$ with the vertical. We need to find the value of $\omega$.

1. Draw an FBD for the mass $m$, showing tension $T$ and gravity $mg$.
2. Resolve tension $T$ into vertical ($T\cos\theta$) and horizontal ($T\sin\theta$) components.
3. Apply Newton's second law:
   • Vertical equilibrium: $T\cos\theta = mg$.
   • Horizontal motion: $T\sin\theta = m\omega^2 r_c$.
4. Determine the radius of the circular path $r_c$. Since the mass hangs from the rim (radius $R$) with a string of length $R$ making an angle $\theta$ with the vertical, the horizontal distance from the axis of rotation to the mass is $R + R\sin\theta = R(1+\sin\theta)$.
5. Substitute $r_c$ into the horizontal motion equation and divide by the vertical equilibrium equation to eliminate $T$.
6. Solve for $\omega$.

The final answer is $\omega = \sqrt{\frac{g \tan\theta}{R(1 + \sin\theta)}}$.