A string 0.5 m long is used to whirl a 1 kg stone in a vertical circle at a uniform velocity of 5 m s-1. What is the tension in the string when the stone is at the top of the circle?

Question

A string 0.5 m long is used to whirl a 1 kg stone in a vertical circle at a uniform velocity of 5 m s-1. What is the tension in the string when the stone is at the top of the circle? (A) 9.8 N (B) 30.4 N (C) 40.2 N (D) 59.8 N

Tardigrade Admin · Accepted Answer

The centripetal force needed to keep the stone moving at 5 m s^-1 is

F_{c} = \frac{m v ^{2}}{r} = \frac{( 1 k g ) ( 5 m s ^{- 1} ) ^{2}}{0.5 m} = 50 N

The weight of the stone is
W = mg = (1 kg) (9.8 m s^-2) = 9.8 N.
At the top of the circle,
T = F_c - W = 50 N - 9.8 N = 40.2 N

Q. A string 0.5m long is used to whirl a 1kg stone in a vertical circle at a uniform velocity of 5ms−1. What is the tension in the string when the stone is at the top of the circle?

9.8 N

30.4 N

40.2 N

59.8 N

The centripetal force needed to keep the stone moving at 5 m s-1 is Fc​=rmv2​=0.5m(1kg)(5ms−1)2​=50N The weight of the stone is W = mg = (1 kg) (9.8 m s-2) = 9.8 N. At the top of the circle, T = Fc - W = 50 N - 9.8 N = 40.2 N

Q. A string $0.5 m$ long is used to whirl a $1 k g$ stone in a vertical circle at a uniform velocity of $5 m s^{- 1}$ . What is the tension in the string when the stone is at the top of the circle?

The centripetal force needed to keep the stone moving at 5 m s^-1 is
$F_{c} = \frac{m v ^{2}}{r} = \frac{( 1 k g ) ( 5 m s ^{- 1} ) ^{2}}{0.5 m} = 50 N$
The weight of the stone is
W = mg = (1 kg) (9.8 m s^-2) = 9.8 N.
At the top of the circle,
T = F_c - W = 50 N - 9.8 N = 40.2 N