Question

Two bodies with moment of inertia I and I' ( I > I' ) have equal angular momentum, If E and E' are the rotational Kinetic energy , then
a) E = E'
b) E > E'
c) E < E'
d) E > or = E'

Answer 1

Two bodies with moment of inertia I and I' (I > I') have equal angular momentum, If E and E' are the rotational Kinetic energy, then

a) $E = E'$

b) E > E'

c) E < E'

d) E > or = E'

Solution:

Angular momentums of the bodies are equal:

$$L_1 = L_2$$

Formula of the angular momentum (I - body's rotational inertia, $\omega$ - rotational velocity):

$$L_m = I \cdot \omega \Rightarrow$$ $$L_1 = I \cdot \omega_1$$ $$L_1 = I' \cdot \omega_2$$

(2) and (3) in (1):

$$I \cdot \omega_1 = I' \cdot \omega_2$$ $$\omega_2 = \omega_1 \cdot \frac{I}{I'}$$

Formula of the rotational Kinetic energy:

$$E_{\text{kinetic}} = \frac{I \cdot \omega^2}{2} = \frac{(I \cdot \omega) \cdot \omega}{2} = \frac{L \cdot \omega}{2} \Rightarrow$$ $$E = \frac{L_1 \cdot \omega_1}{2}$$ $$E' = \frac{L_2 \cdot \omega_2}{2}$$

(4) in (6):

$$E' = \frac{L_2 \cdot \omega_1}{2} \cdot \frac{I}{I'}$$

To find the inequality between the energies, we must subtract $E$ from $E'$

$$E' - E = \frac{L_2 \cdot \omega_1}{2} \cdot \frac{I}{I'} - \frac{L_1 \cdot \omega_1}{2};$$ $$L_1 = L_2 = L \Rightarrow$$ $$E' - E = \frac{L \omega_1}{2} \cdot \left(\frac{I}{I'} - 1\right)$$

If $I > I'$, then $\frac{I}{I'} > 1 \Rightarrow \left(\frac{I}{I'} - 1\right) > 0$.

It means that:

$$E' - E > 0$$ $$E < E'$$

Answer: c) $E < E'$.