Question

A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed. The speed of the stone is 4 m/sec. The tension in the string will be 52 N, when the stone is

• A. At the top of the circle
• B. At the bottom of the circle
• C. Halfway down
• D. None of the above

Answer 1

Answer: (B)

At the bottom of the circle

Answer 2

$m g = 20 N$ and $\frac { m v ^ { 2 } } { r } = \frac { 2 \times ( 4 ) ^ { 2 } } { 1 } = 32 \mathrm {~N}$

It is clear that 52 N tension will be at the bottom of the circle. Because we know that $T _ { \text {Bottom } } = m g + \frac { m v ^ { 2 } } { r }$