Question

A stone of mass '(m)' kg is tied to a string of length '(L)' m and moved in a vertical circle of radius (49 cm) in a vertical plane. If it completes (30) revolutions per minute, the tension in the string when it is at the lowermost point is nearly [Take (\pi^2 = 10) and (g = 10 m/s^2)]

• A. ((90 m) N)
• B. ((60 m) N)
• C. ((45 m) N)
• D. ((15 m) N)

Hint:

Tension is maximum at the bottom ($mg + mr\omega^2$) and minimum at the top ($mr\omega^2 - mg$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A stone performs vertical circular motion. We need to find the tension at the bottom-most point.
Step 2: Key Formula or Approach:
1) Tension at lowest point: \(T = mg + \frac{mv^2}{r} = m(g + \omega^2 r)\).
2) Angular frequency \(\omega = 2\pi f\).
Step 3: Detailed Explanation:
Given:
Radius \(r = 49 \text{ cm} = 0.49 \text{ m}\)
Frequency \(f = 30 \text{ rpm} = \frac{30}{60} \text{ rev/s} = 0.5 \text{ Hz}\)
Angular speed \(\omega = 2 \times \pi \times 0.5 = \pi \text{ rad/s}\)
Acceleration due to gravity \(g = 10 \text{ m/s}^2\)
Using the formula for tension at the bottom:
\[ T = m(g + \omega^2 r) \]
Substitute the values:
\[ T = m(10 + \pi^2 \times 0.49) \]
Since \(\pi^2 \approx 10\):
\[ T = m(10 + 10 \times 0.49) \]
\[ T = m(10 + 4.9) = 14.9 \text{ m N} \]
Rounding to the nearest choice:
\[ T \approx 15 \text{ m N} \]
Step 4: Final Answer:
The tension is nearly \((15 \text{ m})\text{ N}\).