Calculating Kinetic Energy in Circular Motion

Let's determine the kinetic energy of a mass moving along a circular path. The kinetic energy of an object is given by the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the linear velocity. For an object moving in a circular path of radius \(r\) with angular velocity \(\omega\), the linear velocity \(v\) is related by \(v = r\omega\). Therefore, the kinetic energy can also be expressed as \(KE = \frac{1}{2}m(r\omega)^2 = \frac{1}{2}mr^2\omega^2\).

We are given the following information:

• Mass, \(m = 5\) kg
• Radius of the circular path, \(r = 1\) m
• Rate of revolution = 300 revolutions per minute

First, we need to convert the rate of revolution from revolutions per minute to revolutions per second (frequency, \(f\)).

Frequency, \(f = \frac{\text{Number of revolutions}}{\text{Time in seconds}}\)

\(f = \frac{300 \text{ revolutions}}{1 \text{ minute}} = \frac{300 \text{ revolutions}}{60 \text{ seconds}}\)

\(f = 5\) revolutions per second (Hz)

Next, we need to calculate the angular velocity, \(\omega\), from the frequency. The relation between angular velocity and frequency is \(\omega = 2\pi f\).

\(\omega = 2\pi \times 5\) rad/s

\(\omega = 10\pi\) rad/s

Now we can calculate the kinetic energy using the formula \(KE = \frac{1}{2}mr^2\omega^2\).

Substitute the given values into the formula:

\(KE = \frac{1}{2} \times 5 \text{ kg} \times (1 \text{ m})^2 \times (10\pi \text{ rad/s})^2\)

\(KE = \frac{1}{2} \times 5 \times 1 \times (100\pi^2)\)

\(KE = \frac{1}{2} \times 5 \times 100\pi^2\)

\(KE = \frac{1}{2} \times 500\pi^2\)

\(KE = 250\pi^2\) Joules

So, the kinetic energy of the mass is \(250\pi^2\).

Let's compare this result with the given options:

• Option 1: \(100 \pi^2\)
• Option 2: \(0\)
• Option 3: \(250 \pi^2\)
• Option 4: \(6 \pi^2\)

Our calculated kinetic energy matches Option 3.