A 2kg stone tied at the end of a string 1m long is whirled along a vertical circle at a constant speed of 4 ms- 1 . What will be the tension in the string (in newton) at the bottom-most point?

Question

A 2kg stone tied at the end of a string 1m long is whirled along a vertical circle at a constant speed of 4 ms- 1 . What will be the tension in the string (in newton) at the bottom-most point? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Here, m=2 kg,r=1 m,v=4 ms- 1 Tension at the bottom of the circle, TL=mg+(m v2/r) =20+(2 × 42/1)=52 N

Q. A $2 k g$ stone tied at the end of a string $1 m$ long is whirled along a vertical circle at a constant speed of $4 m s^{- 1}$ . What will be the tension in the string (in newton) at the bottom-most point?

Here, $m = 2 k g, r = 1 m, v = 4 m s^{- 1}$
Tension at the bottom of the circle,
$T_{L} = m g + \frac{m v ^{2}}{r}$
$= 20 + \frac{2 \times 4 ^{2}}{1} = 52 N$

Q. A 2kg stone tied at the end of a string 1m long is whirled along a vertical circle at a constant speed of 4ms−1 . What will be the tension in the string (in newton) at the bottom-most point?

Here, m=2kg,r=1m,v=4ms−1 Tension at the bottom of the circle, TL​=mg+rmv2​ =20+12×42​=52N

Q. A $2 k g$ stone tied at the end of a string $1 m$ long is whirled along a vertical circle at a constant speed of $4 m s^{- 1}$ . What will be the tension in the string (in newton) at the bottom-most point?

Here, $m = 2 k g, r = 1 m, v = 4 m s^{- 1}$
Tension at the bottom of the circle,
$T_{L} = m g + \frac{m v ^{2}}{r}$
$= 20 + \frac{2 \times 4 ^{2}}{1} = 52 N$