Question

A pendulum string of length l is moves up to a horizontal position (fig.) and released. What should the minimum strength of the string be to withstand the tension as the pendulum passes through the position of equilibrium? The mass of the pendulum is m.

• A. 3 mg
• B. 4 mg
• C. 5 mg
• D. 6 mg

Answer 1

Answer: (A)

3 mg

Answer 2

The pendulum passes through the equilibrium position moving along the arc of a circle of radius l with a velocity v. At this moment the bob of the pendulum will possess a centripetal acceleration a = $\frac{v^{2}}{\mathcal{l}}$ directed upwards. This acceleration is provided by the joint action of the force of gravity and the tension in the thread (fig.).

By Newton's second law,

F – mg = m $\frac{v^{2}}{\mathcal{l}}$

and hence F = m $\left( g + \frac{v^{2}}{\mathcal{l}} \right)$

The velocity v is determined from the law of conservation of energy and is

v = $\sqrt{2g\mathcal{l}}$

and therefore F = mg + $\frac{mv^{2}}{\mathcal{l}}$ = 3 mg = 3P

where P is the weight of the bob. The thread must be able to support a load equal to three times the weight of the bob.