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
$4\text{ mg}$
To solve this problem, let's analyze the motion of the particle and calculate the tension difference at points B and C.
A particle of mass \(m\) is suspended from a point O by a string of length \(R\). Given a velocity \(u = 3\sqrt{gR}\) at the bottom (point A), it moves in a vertical circle, so we need to determine the tension difference between points B and C.
- Velocity at point B: At point A, the kinetic energy is given by: \(KE_A = \frac{1}{2} m u^2 = \frac{1}{2} m (3\sqrt{gR})^2 = \frac{1}{2} m \cdot 9gR = \frac{9}{2} mgR\).
The particle moves to point B. The change in height (\(\Delta h\)) from A to B is \(R\) (since B is R/4th of the circle above A).
Potential energy change is \(PE = mgR\).
At B, \(KE_B = \frac{9}{2}mgR - mgR = \frac{7}{2}mgR\).
Solve for velocity at B (\(v_B\)): \(\frac{1}{2} m v_B^2 = \frac{7}{2} mgR\).
Thus, \(v_B = \sqrt{7gR}\). - Velocity at point C: The change in height from A to C is \(2R\).
Potential energy at C is \(PE = 2mgR\).
At C, \(KE_C = \frac{9}{2}mgR - 2mgR = \frac{5}{2}mgR\).
So, \(v_C = \sqrt{5gR}\). - Tension at points B and C:
At point B, the centripetal force equation is: \(T_B - mg = \frac{m v_B^2}{R}\).
Substituting for \(v_B\): \(T_B - mg = \frac{m \cdot 7gR}{R} = 7mg\).
Thus, \(T_B = 8mg\).
At point C: \(T_C + mg = \frac{m v_C^2}{R}\).
Substituting for \(v_C\): \(T_C + mg = \frac{m \cdot 5gR}{R} = 5mg\).
Thus, \(T_C = 4mg\). - Difference in tension:
\(\Delta T = T_B - T_C = 8mg - 4mg = 4mg\).
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