A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be

Correct Option: 250π²

To find the kinetic energy of the mass moving along a circular path, we can follow these steps:

Step 1: Identify the given values
Mass ($m$) = 5 kg
Radius ($r$) = 1 m
Rotational frequency ($N$) = 300 revolutions per minute (rpm)

Step 2: Convert the frequency to revolutions per second (Hz)
Since 1 minute = 60 seconds, the frequency ($f$) in revolutions per second is:
$f=\frac{300}{60}=5\text{revolutions per second}$

Step 3: Calculate the angular velocity ($\omega$)
The relation between angular velocity and frequency is:
$\omega =2\pi f$
Substituting the value of $f$:
$\omega =2\times \pi \times 5=10\pi \text{rad/s}$

Step 4: Find the linear velocity ($v$)
The linear velocity of an object moving in a circle of radius $r$ is given by:
$v=r\times \omega$
Substituting $r=1\text{m}$ and $\omega =10\pi \text{rad/s}$:
$v=1\times 10\pi =10\pi \text{m/s}$

Step 5: Calculate the kinetic energy ($\mathrm{K.E.}$)
The formula for kinetic energy is:
$\mathrm{K.E.}=\frac{1}{2}m{v}^2$
Substitute the values of $m$ and $v$ into the equation:
$\mathrm{K.E.}=\frac{1}{2}\times 5\times {\left(10\pi \right)}^2$
Simplify the expression:
$\mathrm{K.E.}=\frac{1}{2}\times 5\times 100{\pi }^2$
$\mathrm{K.E.}=5\times 50{\pi }^2$
$\mathrm{K.E.}=250{\pi }^2\text{J}$

Thus, the kinetic energy of the mass is 250π² J.