Question

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point, is (if $\pi^2 = 9.8$ and $g = 9.8 \, \text{m/s}^2$)

• A. 97 N
• B. 9.8 N
• C. 8.82 N
• D. 17.8 N

Answer 1

Answer: (B)

9.8 N

Answer 2

**Given: **
- Mass of the stone, $m = 900 \, \text{g} = 0.9 \, \text{kg}$
- Radius of the circle, $r = 1 \, \text{m}$
- Angular velocity in rpm, $\omega = 10 \, \text{rpm}$

Step 1. Convert rpm to rad/s:

$\omega = 10 \times \frac{2\pi}{60} = \frac{\pi}{3} \, \text{rad/s}$
]

Step 2. Calculate the centripetal force at the lowest point:
The centripetal force $F_c = m\omega^2r$:

$F_c = 0.9 \times \left(\frac{\pi}{3}\right)^2 \times 1 = 0.9 \times \frac{\pi^2}{9} = 0.9 \times \frac{9.8}{9} = 0.98 \, \text{N}$

Step 3. Calculate the tension $T$ at the lowest point:
At the lowest point, the tension $T$ in the string must support both the gravitational force and the centripetal force. Thus:
$T = mg + F_c$
$T = (0.9 \times 9.8) + 0.98 = 8.82 + 0.98 = 9.8 \, \text{N}$
Thus, the tension in the string at the lowest point is 9.8 N.

The Correct Answer is: 9.8 N