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).

Key concepts

Centripetal Force

Centripetal force is crucial for any object moving in a circular path, as it keeps the object in its path without flying off. This force acts towards the center of the circle. It is not a new kind of force, but rather the net force sum, such as tension, gravity, or friction, that acts to maintain circular motion. In mathematics, centripetal force is expressed by the formula:\[ F_c = mr\omega^2 \]Where:

• \( F_c \) is the centripetal force.
• \( m \) is the mass of the object.
• \( r \) is the radius of the circular path.
• \( \omega \) is the angular velocity.

By understanding this formula, one can analyze how changing the mass, radius, or angular velocity affects the centripetal force needed for an object to maintain its circular path.

Tension Force

In scenarios involving circular motion, such as a stone tied to a string, tension force plays a pivotal role. This force is what keeps the stone moving in a circle, effectively acting as the centripetal force. Tension is the force transmitted through the string, cable or any material that is being pulled from two opposite ends.In uniform circular motion, tension force can be equated to the centripetal force required to maintain the path. Thus:\[ T = F_c = mr\omega^2 \]Here, the tension in the string must be sufficient to provide the necessary centripetal force. If the tension exceeds the breaking point of the string, which in this case is 900 N, the string will snap, and the object will no longer follow the circular path.Understanding the relationship between tension and centripetal force is essential in problems involving an object restrained by a tether in circular motion.

Angular Velocity

Angular velocity, often symbolized by \( \omega \), is a measure of how fast an object rotates or revolves relative to another point. It is a key factor in determining the dynamics of circular motion.Angular velocity is expressed in radians per second (rad/s) and relates to the frequency of rotation and the time it takes for one complete rotation or revolution. In uniform circular motion, it remains constant, meaning the object turns through equal angles in equal times.Using the equation:\[ \omega = \sqrt{\frac{T}{mr}} \]We can determine the maximum angular velocity possible before the tension exceeds the string's breaking limit. This is crucial in systems where material strength is a limiting factor for rotational speed.Keeping track of angular velocity helps in analyzing and designing systems subject to circular motion constraints.

Breaking Tension

Breaking tension is defined as the maximum force a material, such as a string or rope, can withstand before it breaks. In physics problems like the one involving a stone attached to a string, knowing the breaking tension is essential for calculating the maximum parameters of motion, such as angular velocity.Once the force in the string equals or surpasses the breaking tension, which is \( T = 900 \mathrm{~N} \) in this scenario, it will fail, and the object in motion will no longer be held in its circular path.It's essential to:

• Determine the maximum angular velocity by ensuring the calculated centripetal force does not exceed the breaking tension.
• Understand that if the tension in the material surpasses the breaking tension, the material will snap and the motion will cease or change dramatically.

In engineering and safety calculations, understanding breaking tension is crucial for ensuring the durability and reliability of structures and devices undergoing rotational forces.