Question

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with speed 40 rev/min in a horizontal plane. What is the tension in the string and what is the maximum speed with which the stone can be whirled around, if the string can withstand a maximum tension of 200 N?

• A. 6 N, 35 m/s
• B. 6 N, 37 m/s
• C. 7.5 N, 46 m/s
• D. 8 N, 38 m/s

Hint:

For circular motion, the tension in the string is related to the mass of the object, the velocity, and the radius. To find the maximum speed, use the maximum allowable tension.

Answer 1

The Correct Option is A

First, we calculate the speed of the stone using the formula for circular motion.
The linear velocity \( v \) is related to the angular velocity \( \omega \) by the relation: \[ v = r \omega \] where \( r = 1.5 \, \text{m} \) and \( \omega \) is the angular speed.
The number of revolutions per minute is given as 40 rev/min, which we convert to radians per second: \[ \omega = 40 \times \frac{2\pi}{60} = \frac{4\pi}{6} = 2.094 \, \text{rad/s} \] So, the speed of the stone is: \[ v = 1.5 \times 2.094 = 3.141 \, \text{m/s} \] Now, the tension \( T \) in the string is related to the centripetal force by: \[ T = \frac{m v^2}{r} \] Substituting the known values: \[ T = \frac{0.25 \times (3.141)^2}{1.5} = 6 \, \text{N} \] Next, we find the maximum speed \( v_{\text{max}} \) at which the stone can be whirled around, given the maximum tension of 200 N.
Using the formula for tension: \[ T_{\text{max}} = \frac{m v_{\text{max}}^2}{r} \] Solving for \( v_{\text{max}} \): \[ 200 = \frac{0.25 \times v_{\text{max}}^2}{1.5} \] \[ v_{\text{max}}^2 = \frac{200 \times 1.5}{0.25} = 1200 \] \[ v_{\text{max}} = \sqrt{1200} = 34.64 \, \text{m/s} \]
Thus, the correct answer is (a).