Question 18: Work, Energy and Power & Circular Motion

Question

A body is moving along a circular path of radius '$$r$$' with a frequency of revolution numerically equal to the radius of the circular path. What is the acceleration of the body if radius of the path is $$\left(\frac{5}{\pi}\right) m$$ ?

Accepted Answer

First, let's analyze the given information. The radius of the circular path, $$r$$, is given as $$\frac{5}{\pi}$$ meters. The frequency of revolution, $$f$$, is numerically equal to the radius, so $$f = r = \frac{5}{\pi}$$ Hz.
The acceleration we need to find is the centripetal acceleration, which is given by the formula:
$$a = \omega^2 r$$
Here, $$\omega$$ (omega) is the angular velocity, which can be calculated from the frequency $$f$$ using the relationship:
$$\omega = 2 \pi f$$
Substitute $$f = \frac{5}{\pi}$$ into the equation for $$\omega$$:
$$\omega = 2 \pi \left(\frac{5}{\pi}ight) = 10 \, 	ext{rad/s}$$
Now, substitute $$\omega = 10 \, 	ext{rad/s}$$ and $$r = \frac{5}{\pi}$$ meters into the centripetal acceleration formula:
$$a = \omega^2 r = (10)^2 \left(\frac{5}{\pi}ight)$$
Thus, the acceleration $$a$$ is:
$$a = 100 \left(\frac{5}{\pi}ight) = \frac{500}{\pi} \, 	ext{m/s}^2$$
Therefore, the correct option is:
Option D: $$\left(\frac{500}{\pi}ight) \mathrm{ms}^{-2}$$

Answer

$$
100 \pi \mathrm{~ms}^{-2}
$$

Answer

$$
500 \pi \mathrm{~ms}^{-2}
$$

Answer

$$
25 \pi \mathrm{~ms}^{-2}
$$

Answer

$$
\left(\frac{500}{\pi}\right) \mathrm{ms}^{-2}
$$

Work, Energy and Power & Circular Motion

A body is moving along a circular path of radius '$r' with a frequency of revolution numerically equal to the radius of the circular path. What is the acceleration of the body if radius of the path is \left(\frac{5}{\pi}\right) m$ ?

Choose the correct answer: