Question

find the kinetic energy of a rigid body rotating about a fixed point.

Answer 1

Answer on Question 58586, Physics, Mechanics, Relativity

Question:

Find the kinetic energy of a rigid body rotating about a fixed point.

Solution:

A rigid body rotating with uniform angular speed $\omega$ about a fixed point possesses kinetic energy of rotation. We can calculate its value by summing up the individual kinetic energies of all the particles of which the body is composed. A particle of mass $m_{1}$ located at distance $r_1$ from the axis of rotation has kinetic energy given by $\frac{1}{2} m_1 v_1^2$, here $v_{1}$ is the speed of the particle. Then, we can write the formula for the total kinetic energy:

$$E _ {k} = \frac {1}{2} m _ {1} v _ {1} ^ {2} + \frac {1}{2} m _ {2} v _ {2} ^ {2} + \dots + \frac {1}{2} m _ {n} v _ {n} ^ {2} = \sum_ {i = 1} ^ {n} \frac {1}{2} m _ {i} v _ {i} ^ {2},$$

Each particle of a rigid body rotates with uniform angular speed $\omega$. Then, using the relation between linear and angular variables $(v = \omega r)$ and substituting it into the previous equation, we get:

$$E _ {k} = \frac {1}{2} m _ {1} r _ {1} ^ {2} \omega^ {2} + \frac {1}{2} m _ {2} r _ {2} ^ {2} \omega^ {2} + \dots + \frac {1}{2} m _ {n} r _ {n} ^ {2} \omega^ {2} = \frac {1}{2} \omega^ {2} (m _ {1} r _ {1} ^ {2} + m _ {2} r _ {2} ^ {2} + \dots m _ {n} r _ {n} ^ {2}).$$

Let's denote the factor in parentheses by the letter $I$ (here, $I$ is the moment of inertia of the rotating body with respect to the particular axis of rotation):

$$I = m _ {1} r _ {1} ^ {2} + m _ {2} r _ {2} ^ {2} + \dots m _ {n} r _ {n} ^ {2} = \sum_ {i = 1} ^ {n} m _ {i} r _ {i} ^ {2}.$$

Finally, we can write the kinetic energy of a rigid body rotating about a fixed point as:

$$E _ {k} = \frac {1}{2} I \omega^ {2}.$$

Answer:

$$E _ {k} = \frac {1}{2} I \omega^ {2}.$$

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