Question:
A body of mass \(0.6\text{ kg}\) is moving along a circular path of radius \(1\text{ m}\). If the body moves with
\[
\frac{900}{\pi}
\]
revolutions per minute, its kinetic energy is
A body of mass \(0.6\text{ kg}\) is moving along a circular path of radius \(1\text{ m}\). If the body moves with
\[
\frac{900}{\pi}
\]
revolutions per minute, its kinetic energy is
Show Hint
Remember:
\[
\omega=2\pi f
\]
and
\[
v=\omega r
\]
for circular motion problems.
Updated On: Jun 25, 2026
- \(120\text{ J}\)
- \(270\text{ J}\)
- \(360\text{ J}\)
- \(240\text{ J}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Convert revolutions per minute into angular speed.
Given frequency: \[ n=\frac{900}{\pi}\text{ revolutions per minute} \] Frequency per second is \[ f=\frac{900}{\pi\times 60} \] \[ f=\frac{15}{\pi}\text{ revolutions per second} \] Angular speed: \[ \omega=2\pi f \] \[ \omega=2\pi\left(\frac{15}{\pi}\right) \] \[ \omega=30\text{ rad s}^{-1} \]
Step 2: Find the linear speed.
Using \[ v=\omega r \] Given \[ r=1\text{ m} \] So, \[ v=30\times 1 \] \[ v=30\text{ m s}^{-1} \]
Step 3: Calculate kinetic energy.
Kinetic energy is \[ K=\frac12 mv^2 \] Substitute: \[ K=\frac12(0.6)(30)^2 \] \[ K=0.3\times 900 \] \[ K=270\text{ J} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{270\text{ J}} \]
Given frequency: \[ n=\frac{900}{\pi}\text{ revolutions per minute} \] Frequency per second is \[ f=\frac{900}{\pi\times 60} \] \[ f=\frac{15}{\pi}\text{ revolutions per second} \] Angular speed: \[ \omega=2\pi f \] \[ \omega=2\pi\left(\frac{15}{\pi}\right) \] \[ \omega=30\text{ rad s}^{-1} \]
Step 2: Find the linear speed.
Using \[ v=\omega r \] Given \[ r=1\text{ m} \] So, \[ v=30\times 1 \] \[ v=30\text{ m s}^{-1} \]
Step 3: Calculate kinetic energy.
Kinetic energy is \[ K=\frac12 mv^2 \] Substitute: \[ K=\frac12(0.6)(30)^2 \] \[ K=0.3\times 900 \] \[ K=270\text{ J} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{270\text{ J}} \]
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