Problem 34 - Master Resource Book in JEE Main Physics

Question

A stone tied to the end of a string \(80 \mathrm{~cm}\) long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in \(25 \mathrm{~s}\), what is the magnitude and direction of acceleration of the stone? (a) \(9.9 \mathrm{~m} / \mathrm{s}^{2}\) along the tangent (b) \(7.9 \mathrm{~m} / \mathrm{s}^{2}\) along the radius (c) \(9.9 \mathrm{~m} / \mathrm{s}^{2}\) along the radius (d) None of the above

Step-by-Step Solution

Answer

(c) \(9.9 \mathrm{~m/s}^2\) along the radius.

Step 1: Determine the Radius in Meters

The length of the string is given as \( 80 \text{ cm} \). First, convert it to meters because SI units are used in calculations:\[ r = 80 \text{ cm} = 0.8 \text{ m} \]

Step 2: Calculate the Angular Velocity

The stone makes 14 revolutions in 25 seconds. First, find the frequency:\[ f = \frac{14}{25} \text{ rev/s} \]The angular velocity \( \omega \) is given by:\[ \omega = 2\pi f = 2\pi \times \frac{14}{25} \approx 3.52 \text{ rad/s} \]

Step 3: Calculate the Centripetal Acceleration

The centripetal acceleration \( a_c \) is calculated using the formula: \[ a_c = \omega^2 \times r \]Substitute the values we have:\[ a_c = (3.52)^2 \times 0.8 \approx 9.9 \text{ m/s}^2 \]

Step 4: Determine the Direction of Acceleration

In circular motion, the acceleration is always directed towards the center of the circle along the radius. Therefore, the direction of the acceleration is radial.

Key Concepts

Circular Motion

Circular motion refers to the movement of an object along a circular path. This type of motion is quite common and can be observed in many everyday phenomena, like spinning a ball on a string or the rotation of the Earth around the Sun. In circular motion, the object revolves around a fixed point, which is called the center of the circle. Unlike linear motion, where an object moves from one point to another, circular motion describes a movement that is repetitive and follows a curved path.
It is important to note that even if the speed of the object remains constant in circular motion, its velocity is changing constantly. This is because velocity is a vector quantity, meaning it has both magnitude and direction. Since the direction is always changing as the object moves around the circle, the velocity changes as well.

There are various forms of circular motion including uniform circular motion, where the speed is constant, and non-uniform, where the speed can vary. The concept of centripetal force, which we will discuss later, plays a crucial role in maintaining circular motion, ensuring the object does not move off the curved path.

Angular Velocity

Angular velocity, denoted typically by the Greek letter \( \omega \), is a measure of the rate of rotation. It describes how quickly an object rotates or revolves around a point, which is the center of the circular path. In simpler terms, angular velocity tells us how fast something is spinning.
The formula for angular velocity is:

• \( \omega = \frac{\Delta \theta}{\Delta t} \)

where \( \Delta \theta \) is the change in angular position (in radians) and \( \Delta t \) is the change in time. For an object moving with uniform circular motion, angular velocity is constant.
In our specific problem, we calculated the angular velocity by first determining the frequency of the stone's revolutions. Frequency, measured in revolutions per second (rev/s), is then used to find the angular velocity using the formula \( \omega = 2\pi f \). If we plug our values into the formula, we arrive at an angular velocity of approximately 3.52 rad/s. This angular velocity indicates how quickly the stone is revolving around the circle's center.

Centripetal Force

Centripetal force is a critical concept in understanding circular motion. It is the inward force required to keep an object moving in a circular path. Without centripetal force, the object would move off in a straight line due to inertia, as per Newton's first law of motion.
The magnitude of the centripetal force \( F_c \) needed to keep an object in circular motion is given by the equation:

• \( F_c = m \omega^2 r \)

where:

• \( m \) is the mass of the object,
• \( \omega \) is the angular velocity, and
• \( r \) is the radius of the circular path.

Centripetal force acts perpendicular to the velocity of the object and points towards the center of the circle, thus continuously changing the direction of the velocity.
This force maintains the circular path of the stone tied to the string in our exercise, pulling it inward as it spins. While we did not directly calculate the force in our problem, we found the centripetal acceleration, which is a component often used alongside centripetal force in motion analysis. The stone's centripetal acceleration tells us how much the velocity of the stone is changing as it moves along the circular path. In simpler terms, it indicates how tightly the stone adheres to its circular path.