Q138 / 180 · COMEDK 2024 (Morning Shift) · Physics
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 $\left(\dfrac{5}{\pi}\right) m$ ?
- A$500 \pi \mathrm{~ms}^{-2}$
- B$25 \pi \mathrm{~ms}^{-2}$
- C$100 \pi \mathrm{~ms}^{-2}$
- Correct answerD$\left(\dfrac{500}{\pi}\right) \mathrm{ms}^{-2}$
Solution
The acceleration of a body moving in a circular path is the centripetal acceleration, given by $a = \omega^2 r$.
The angular velocity $\omega$ is related to the frequency $f$ by the formula $\omega = 2 \pi f$.
Given that the frequency $f$ is numerically equal to the radius $r$, we have $f = r$.
Substituting $f = r$ into the expression for $\omega$, we get $\omega = 2 \pi r$.
The acceleration $a$ becomes $a = (2 \pi r)^2 r = 4 \pi^2 r^3$.
Given the radius $r = \dfrac{5}{\pi}$ m, we substitute this value into the expression for $a$:
$a = 4 \pi^2 \left( \dfrac{5}{\pi} \right)^3$
$a = 4 \pi^2 \left( \dfrac{125}{\pi^3} \right)$
$a = \dfrac{4 \times 125}{\pi} = \dfrac{500}{\pi} \text{ ms}^{-2}$.
Answer: $\left(\dfrac{500}{\pi}\right) \mathrm{ms}^{-2}$
The angular velocity $\omega$ is related to the frequency $f$ by the formula $\omega = 2 \pi f$.
Given that the frequency $f$ is numerically equal to the radius $r$, we have $f = r$.
Substituting $f = r$ into the expression for $\omega$, we get $\omega = 2 \pi r$.
The acceleration $a$ becomes $a = (2 \pi r)^2 r = 4 \pi^2 r^3$.
Given the radius $r = \dfrac{5}{\pi}$ m, we substitute this value into the expression for $a$:
$a = 4 \pi^2 \left( \dfrac{5}{\pi} \right)^3$
$a = 4 \pi^2 \left( \dfrac{125}{\pi^3} \right)$
$a = \dfrac{4 \times 125}{\pi} = \dfrac{500}{\pi} \text{ ms}^{-2}$.
Answer: $\left(\dfrac{500}{\pi}\right) \mathrm{ms}^{-2}$
Hence, the correct answer is option: 'D'.