Question

A particle of mass 2 kg is tied to a string of length 1 m is performing circular motion with the frequency 120 rpm.

Find the rotational Kinetic Energy of the particle.​

Answer 1

Answer:

• Mass (M) = 2 kg
• Length / Radius (R) = 1 m
• Frequency = 120 rpm = 2 rps
• Rotational kinetic energy = ?

First we need to calculate the moment of inertia :

$\longrightarrow\:\rm I = MR^2$

$\longrightarrow\:\rm I = 2 \times (1)^2$

$\longrightarrow\: \underline{ \underline{\rm I = 2 \: kg. {m}^{2}}}$

After finding moment of inertia we need to calculate the angular velocity of the particle

$\longrightarrow\:\rm \omega = 2 \pi n$

$\longrightarrow\:\rm \omega = 2 \times 3.14 \times \dfrac{120}{60}$

$\longrightarrow\:\rm \omega = 2 \times 3.14 \times 2$

$\longrightarrow\: \underline{ \underline{\rm \omega = 12.56 \: {rad}^{ - 1}}}$

Now, let's find the rotational kinetic energy of the particle :

$\longrightarrow\:\rm K.E = \dfrac{1}{2}\times I \times \omega^2$

$\longrightarrow\:\rm K.E = \dfrac{1}{2}\times 2 \times (12.56)^2$

$\longrightarrow\:\rm K.E =(12.56)^2$

$\longrightarrow\: \underline{ \underline{\rm K.E =157.75 \:J }}$