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

Answer 1

Answer: (C)

40.2 N

Answer 2

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