Question

A ball is held at rest at position A in Fig. P5.115 by two light strings. The horizontal string is cut, and the ball starts swinging as a pendulum. Position B is the farthest to the right that the ball can go as it swings back and forth. What is the ratio of the tension in the supporting string at B to its value at A before the string was cut?

Short answer

The required ratio is $\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}={\cos }^2\mathrm{\beta }$.

Step-by-step solution

Concept/Significance of particle under a net force:

If a particle of mass (m) experience a nonzero net force (F), its acceleration (a) is related to the net force by Newton’s second law.

$\mathbf{\sum }_{\mathbf{}}\mathbf{F}\mathbf{=}\mathbf{ma}$

Find the ratio of the tension in the supporting string at B to its value at A before the string was cut:

Consider the following free-body diagram of ball A.

At point A the sting is in equilibrium so according to Newton’s second law the force is each direction is equal to zero.

$$\begin{gathered} \sum F_y=0 \\ {T}_{A}cos\beta =mg \end{gathered}$$

${\mathrm{T}}_{\mathrm{A}}=\frac{\mathrm{mg}}{\mathrm{cos\beta }}$ ….. (1)

After the rope is cut the ball will move in simple harmonic motion and one type of simple harmonic motion is the uniform circular motion so focus in radial direction were the tension force equalizes the weight, so at point B the ball is not in equilibrium but the acceleration is zero in radial direction so the component of weight is in equilibrium with tension at (b).

Consider the following free-body diagram of ball B.

Apply the Newton’s second law the force.

$\sum F_R=0$

$T_B=mgcos\beta$ ….. (2)

Divide equations (1) and (2) as below.

$$\begin{gathered} \frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}=\frac{\mathrm{mgcos\beta }}{{\displaystyle \frac{\mathrm{mg}}{\mathrm{cos\beta }}}} \\ ={\cos }^2\mathrm{\beta } \end{gathered}$$

Therefore, the required ratio is $\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}={\cos }^2\mathrm{\beta }$.