Question

A particle of mass \( m \) is suspended from a point O by a string of length \( R \). It is given a velocity \( u = 3\sqrt{gR} \) at the bottom. The difference in tension at point B and at the point C is

• A. \( 6\text{ mg} \)
• B. \( 4\text{ mg} \)
• C. \( 3\text{ mg} \)
• D. \( 8\text{ mg} \)

Hint:

For vertical circular motion starting with a bottom velocity \( u \), the tension at any angle \( \theta \) with the vertical is given by \( T = \frac{m v^2}{R} + mg \cos\theta \). By expressing velocity \( v \) in terms of \( u \) using energy conservation, we obtain:
\[ T(\theta) = \frac{m u^2}{R} - 2mg + 3mg \cos\theta \]
- At \( B \) (\( \theta = 90^\circ \)): \( T_B = \frac{mu^2}{R} - 2mg = 9mg - 2mg = 7mg \).
- At \( C \) (\( \theta = 180^\circ \)): \( T_C = \frac{mu^2}{R} - 2mg - 3mg = 9mg - 5mg = 4mg \).
This general formula \( T(\theta) = \frac{mu^2}{R} - 2mg + 3mg\cos\theta \) can help you solve any tension question instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A particle of mass \( m \) undergoes vertical circular motion of radius \( R \) with an initial velocity \( u = 3\sqrt{gR} \) at the bottommost point \( A \). We need to determine the difference in the tension of the string when the particle is at the horizontal position \( B \) and at the highest position \( C \).

Step 2: Key Formula or Approach:
1. Conservation of Mechanical Energy:
\[ v^2 = u^2 - 2gh \]
where \( h \) is the vertical height from the bottommost point.
2. Centripetal Force Equations:
- At the horizontal position \( B \) (\( h_B = R \)):
\[ T_B = \frac{m v_B^2}{R} \]
- At the highest point \( C \) (\( h_C = 2R \)):
\[ T_C + mg = \frac{m v_C^2}{R} \]

Step 3: Detailed Explanation:
1. Analysis at point B (height \( h_B = R \)):
Using energy conservation:
\[ v_B^2 = u^2 - 2gR = (3\sqrt{gR})^2 - 2gR = 9gR - 2gR = 7gR \]
The tension \( T_B \) provides the entire centripetal acceleration at this position:
\[ T_B = \frac{m v_B^2}{R} = \frac{m(7gR)}{R} = 7mg \]
2. Analysis at point C (height \( h_C = 2R \)):
Using energy conservation:
\[ v_C^2 = u^2 - 2g(2R) = 9gR - 4gR = 5gR \]
At the top, both tension \( T_C \) and gravity \( mg \) act towards the center:
\[ T_C + mg = \frac{m v_C^2}{R} = \frac{m(5gR)}{R} = 5mg \]
\[ T_C = 5mg - mg = 4mg \]
3. Calculating the difference in tension:
\[ T_B - T_C = 7mg - 4mg = 3mg \]

Step 4: Final Answer:
The difference in tension between point B and point C is \( 3\text{ mg} \).