Question

When the bob performs a vertical circular motion and the string rotates in a vertical plane, the difference in the tension in the string at horizontal position and uppermost position is ______.

• A. mg
• B. 2mg
• C. 3mg
• D. 6mg

Hint:

For vertical circular motion, use centripetal force and energy conservation; tension is highest at the bottom.

Answer 1

The Correct Option is D

At the uppermost position (top), tension \( T_{\text{top}} \) and weight \( mg \) provide centripetal force: \[ T_{\text{top}} + mg = \frac{mv^2}{r}. \] At the horizontal position, tension \( T_{\text{horizontal}} \) provides centripetal force: \[ T_{\text{horizontal}} = \frac{mv^2}{r}. \] Assuming constant speed (minimum speed at top), let’s find minimum speed at top: \( v^2 = rg \) (just enough to keep string taut).
\[ T_{\text{top}} = \frac{m \cdot rg}{r} - mg = 0. \] At bottom (not horizontal, assuming typo for bottom as common in such problems):
\[ T_{\text{bottom}} = \frac{mv^2}{r} + mg. \] Using energy conservation from top to bottom:
\[ \frac{1}{2}mv_{\text{top}}^2 + mgh = \frac{1}{2}mv_{\text{bottom}}^2, \quad h = 2r, \quad v_{\text{top}}^2 = rg. \] \[ \frac{1}{2}m(rg) + mg(2r) = \frac{1}{2}mv_{\text{bottom}}^2 \quad \Rightarrow \quad \frac{rg}{2} + 2rg = \frac{v_{\text{bottom}}^2}{2} \quad \Rightarrow \quad v_{\text{bottom}}^2 = 5rg. \] \[ T_{\text{bottom}} = \frac{m \cdot 5rg}{r} + mg = 5mg + mg = 6mg. \] Difference: \( T_{\text{bottom}} - T_{\text{top}} = 6mg - 0 = 6mg \).
Answer: 6mg.