Question 283: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

To find the acceleration of the particle performing uniform circular motion (U.C.M.), we will use the formula for centripetal acceleration:
$$a = \omega^2 r$$
where $$\omega$$ is the angular velocity and $$r$$ is the radius of the circle.
First, we need to determine the angular velocity $$\omega$$. The formula for angular velocity in terms of frequency $$f$$ is:
$$\omega = 2\pi f$$
Given the frequency $$f = 2 \mathrm{~Hz}$$, we have:
$$\omega = 2\pi 	imes 2 = 4\pi \mathrm{~rad/s}$$
Next, we need to find the radius $$r$$ of the circle. Given the diameter as $$50 \mathrm{~cm}$$, the radius $$r$$ is:
$$r = \frac{50 \mathrm{~cm}}{2} = 25 \mathrm{~cm} = 0.25 \mathrm{~m}$$
Now we can calculate the centripetal acceleration:
$$a = \omega^2 r = (4\pi)^2 	imes 0.25 = 16\pi^2 	imes 0.25 = 4\pi^2 \mathrm{~m/s}^2$$
Therefore, the acceleration of the particle is:
Option B
$$4 \pi^2 \mathrm{~m/s}^2$$

Answer

$$2 \pi^2$$

Answer

$$4 \pi^2$$

Answer

$$8 \pi^2$$

Answer

$$\pi^2$$

Work, Energy and Power & Circular Motion

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

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