Question

A string of length $L$ fixed at one end carries a body of mass $m$ at the other end. The mass is revolved in a circle in horizontal plane making angle $\theta$ with vertical. The angular frequency of the body is $\omega$. The tension in the string is

• A. $mL^2\omega$
• B. $mL\omega^2$
• C. $\dfrac{\omega^2}{mL}$
• D. $\dfrac{m\omega^2}{L}$

Hint:

Physics Tip: For conical pendulum: $T\sin\theta=m\omega^2r$ and $T\cos\theta=mg$.

Answer 1

The Correct Option is B

Step 1: Recognize the motion. This is a conical pendulum. The mass moves in a horizontal circle while the string makes constant angle $\theta$ with vertical.

Step 2: Forces acting on mass. Two forces act:
• Weight $mg$ downward
• Tension $T$ along string Horizontal component of tension provides centripetal force.

Step 3: Radius of circular path. If string length is $L$: $$ r=L\sin\theta $$

Step 4: Centripetal force equation. Required centripetal force: $$ T\sin\theta=m\omega^2r $$ Substitute $r=L\sin\theta$: $$ T\sin\theta=m\omega^2(L\sin\theta) $$

Step 5: Simplify. Cancel $\sin\theta$ from both sides: $$ T=mL\omega^2 $$ Thus tension is independent of angle $\theta$ in this form. $$ \therefore \text{Correct option is (B).} $$