Problem 42 - Master Resource Book in JEE Main Physics

Question

A particle moves in a circle of radius \(5 \mathrm{~cm}\) with constant speed and time period \(0.2 \pi \mathrm{s}\). The acceleration of the particle is (a) \(5 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(15 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(25 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(36 \mathrm{~m} / \mathrm{s}^{2}\)

Step-by-Step Solution

Answer

The acceleration of the particle is \(5 \mathrm{~m} / \mathrm{s}^{2}\).

Step 1: Identify the Known Values

We know that the radius of the circle is given as \( r = 5 \text{ cm} = 0.05 \text{ m} \) and the time period of the particle is \( T = 0.2 \pi \text{ s} \).

Step 2: Use the Formula for Centripetal Acceleration

The formula for the centripetal acceleration \( a_c \) of a particle moving in a circle is given by \( a_c = \frac{v^2}{r}\), where \( v \) is the speed of the particle.

Step 3: Relate Speed and Time Period

To find \( v \), the speed, use the relationship between speed, radius, and time period: \( v = \frac{2\pi r}{T} \). Substitute the known values to calculate \( v \): \ \(v = \frac{2\pi \times 0.05}{0.2 \pi} = 0.5\text{ m/s}\)

Step 4: Calculate the Centripetal Acceleration

Substitute \( v = 0.5 \text{ m/s} \) and \( r = 0.05 \text{ m} \) into the centripetal acceleration formula to find \( a_c \): \\(a_c = \frac{0.5^2}{0.05} = 5 \text{ m/s}^2\)

Step 5: Verify with the Given Options

The calculated centripetal acceleration is \( 5 \text{ m/s}^2 \), which corresponds to option (a) in the given choices.

Key Concepts

Uniform Circular Motion

Imagine a particle moving around the edge of a circle. If it moves at a constant speed, we say it's undergoing uniform circular motion. This type of motion is quite common, seen in examples like the moon orbiting the Earth, or a car going around a roundabout at constant speed. Even though the speed of the particle is constant, it continuously changes direction to stay on the circular path.

Here are some key points about uniform circular motion:

• The speed stays the same, but the velocity, which includes direction, does change.
• Circular motion always has a center-point that the object is moving around.
• There is an inward force called centripetal force that keeps objects moving in a circle.

Time Period

In the context of circular motion, the time period is crucial. It tells us how long it takes for one complete revolution or orbit around the circle. Imagine if you time how long it takes for a ball tied to a string and swung around, to return to the same spot—the time you're measuring is the time period.

The time period is closely related to the frequency of motion, which is how many revolutions happen per second. Here's more to know about time period:

• It's denoted usually by the letter \(T\).
• The formula to find the time period when you know the speed and radius is \(T = \frac{2\pi r}{v}\).
• A smaller time period means faster revolutions.

Speed Calculation

Speed calculation in circular motion is an interesting task. It gives us the idea of how fast the object is moving along the circle, though not how fast it is leaving the circle, since it does not leave. The equation linking speed, radius, and time period is incredibly handy.

For uniform circular motion, the speed \(v\) can be calculated as: \(v = \frac{2\pi r}{T}\). Here’s a breakdown of the key steps:

• \(r\) is the circle's radius, providing how far out from the center the particle travels.
• \(T\) is the time period, showing how long one full circle journey takes.
• This formula captures how the larger the radius, the higher the speed for the same time period.