Question

A particle is performing U.C.M. along the circumference of a circle of diameter $50 \text{ cm}$ with frequency $2 \text{ Hz}$. The acceleration of the particle in $\text{m/s}^2$ is

• A. $2\pi^2$
• B. $4\pi^2$
• C. $8\pi^2$
• D. $\pi^2$

Hint:

You can use the combined linear formula shortcut for centripetal acceleration: $a = 4\pi^2 f^2 r$. Plugging in the values directly: $4\pi^2 \cdot (2)^2 \cdot (0.25) = 4\pi^2 \cdot 4 \cdot 0.25 = 4\pi^2 \text{ m/s}^2$ in a single step.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the magnitude of the centripetal acceleration of a particle moving in Uniform Circular Motion (U.C.M.), given the diameter of its circular path and its rotational frequency.

Step 2: Key Formula or Approach:
The centripetal acceleration $a$ of a particle in U.C.M. can be calculated using the formula:
$$a = \omega^2 r$$ where $\omega$ is the angular velocity and $r$ is the radius of the circular path.
The angular velocity is related to frequency $f$ by:
$$\omega = 2\pi f$$

Step 3: Detailed Explanation:
Let's list the given parameters and convert them into standard SI units:
Diameter $= 50 \text{ cm} \implies \text{Radius, } r = 25 \text{ cm} = 0.25 \text{ m} = \frac{1}{4} \text{ m}$
Frequency, $f = 2 \text{ Hz}$
First, calculate the angular velocity $\omega$:
$$\omega = 2\pi (2) = 4\pi \text{ rad/s}$$ Next, substitute $\omega$ and $r$ into the centripetal acceleration formula:
$$a = (4\pi)^2 \cdot (0.25)$$ $$a = 16\pi^2 \cdot \frac{1}{4} = 4\pi^2 \text{ m/s}^2$$

Step 4: Final Answer:
The acceleration of the particle is $4\pi^2 \text{ m/s}^2$, which matches option (B).