Question

A bob of mass m is suspended by a light string of length l it is imparted a horizontal

Answer 1

Complete question:

A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v₀ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming limp only on reaching the topmost point, C. This is shown in Figure. Obtain an expression for (i) v₀; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies ($K_B/K_C$) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.

Answer:

(i) v₀ = √(5gL)

(ii) The speeds at points B and C is √(3gL) and √(gL).

(iii) $K_B/K_C = 3/1$

Explanation:

(i)

Let the velocity of the particle at top most point be v m/s

The particle is in vertical motion.

The centripetal acceleration is given as:

mg = mv²/L

The speed at the point C when sting limps is ⇒ u = √(gL)

On applying energy conservation between A and C, we get,

$\frac{m v_{0}^{2}}{2}+0=\frac{m u^{2}}{2}+m g(2 L)$

$\frac{m v_{0}^{2}}{2}=\frac{m u^{2}}{2}+m g(2 L)$

∴ v₀ = √(5gL)

(ii)

On applying energy conservation on point C, we get,

$\frac{m v_{0}^{2}}{2}+0=\frac{m u_{B}^{2}}{2}+m g(L)$

$\frac{m v_{0}^{2}}{2}=\frac{m u_{B}^{2}}{2}+m g(L)$

The speed at point B is:

∴ $u_B$ = √(3gL)

(iii)

The ratio of the kinetic energy be:

$\frac{K E_{B}}{K E_{C}}=\frac{\frac{m u_{B}^{2}}{2}}{\frac{m u^{2}}{2}}$

$\frac{K E_{B}}{K E_{C}}=\frac{u_{B}^{2}}{u^{2}}$

$\frac{K E_{B}}{K E_{C}}=\frac{2 g L}{g L}$

$\therefore \frac{K E_{B}}{K E_{C}}=\frac{3}{1}$

Answer 2

Answer:

(i):- root 5gL

(ii):- VB=root 3gL

(iii):- 3/1