Question

A stone of mass \(2 \mathrm{~kg}\) is tied to a string of length $0.5 \mathrm{~m}\( It the breaking tension of the string is \)900 \mathrm{~N}$, then the maximum angular velocity the stone can have in uniform circular motion is (A) \(30 \mathrm{rad} / \mathrm{s}\) (B) \(20 \mathrm{rad} / \mathrm{s}\) (C) \(10 \mathrm{rad} / \mathrm{s}\) (D) \(25 \mathrm{rad} / \mathrm{s}\)

Short answer

The maximum angular velocity the stone can have in uniform circular motion is \(30~\frac{rad}{s}\). The correct answer is (A).

Step-by-step solution

Identifying the given information

The given information is as follows: 1. Mass of the stone (m) = \(2~kg\) 2. Length of the string (r) = \(0.5~m\) 3. Breaking tension of the string (T) = \(900~N\) The objective is to find the maximum angular velocity (ω) of the stone in uniform circular motion.

Understanding the required force for uniform circular motion

For a body in uniform circular motion, there should be a centripetal force acting on it. Centripetal force can be calculated using the following formula: \[F_c = mrω^2\] Where: - \(F_c\) is the centripetal force, - m is the mass of the stone, - r is the radius of the circular path, and - ω is the angular velocity.

Relating tension force to centripetal force

The tension force in the string directly provides the centripetal force needed for uniform circular motion. Thus, we can equate the tension force (T) to the centripetal force (Fc): \[T = F_c = mrω^2\] Using the given values for T, m, and r, we can solve for the maximum angular velocity (ω) of the stone.

Solve for the maximum angular velocity

Plug in the given values into the equation derived in Step 3 and solve for ω: \[900~N = (2~kg)(0.5~m)ω^2\] Divide both sides of the equation by (2 kg)(0.5 m) and take the square root: \[ω = \sqrt{\frac{900~N}{(2~kg)(0.5~m)}}\] \[ω = \sqrt{\frac{900}{1}}\] \[ω = 30~\frac{rad}{s}\] So the maximum angular velocity the stone can have in uniform circular motion is \(30~\frac{rad}{s}\). The correct answer is (A).