Question

021 (part 1 of 2) 10.0 points

Answer 1

Answers:

a) $1.17 rad/s$

b) $1.31 J$

Explanation:

a) We can solve this knowing the angular momentum $L$ is conserved, then:

$L_{o}=L_{f}$

$I_{o} \omega_{o}=I_{f} \omega_{f}$ (1)

Where:

$I_{o}$ is the initial moment of inertia of the system

$\omega_{o}=0.69 rad/s$ is the initial angular velocity

$I_{f}$ is the final moment of inertia of the system

$\omega_{f}$ is the final angular velocity

But first, we have to find $I_{o}$ and $I_{f}$:

$I_{o}=I_{s}+2mr_{o}^{2}$ (2)

$I_{f}=I_{s}+2mr_{f}^{2}$ (3)

Where:

$I_{s}=4 kgm^{2}$ is the student's moment of inertia

$m=4 kg$ is the mass of each object

$r_{o}=0.7 m$ is the initial radius

$r_{f}=0.29 m$ is the final radius

Then:

$I_{o}=4 kgm^{2}+2(4 kg)(0.7 m)^{2}=7.92 kgm^{2}$ (4)

$I_{f}=4 kgm^{2}+2(4 kg)(0.29 m)^{2}=4.67 kgm^{2}$ (5)

Substituting the results of (4) and (5) in (1):

$(7.92 kgm^{2}) (0.69 rad/s)=4.67 kgm^{2}\omega_{f}$ (6)

Finding $\omega_{f}$:

$\omega_{f}=1.17 rad/s$ (7) This is the final angular velocity

b) The rotational kinetic energy is:

$K=\frac{1}{2}I \omega^{2}$ (8)

The change in kinetic energy is:

$\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{o} \omega_{o}^{2}$ (9)

Since we already calculated these values, we can solve (9):

$\Delta K=\frac{1}{2}(4.67 kgm^{2}) (1.17 rad/s)^{2}-\frac{1}{2}(7.92 kgm^{2}) (0.69 rad/s)^{2}$ (10)

Finally:

$\Delta K=1.31 J$