Question

PART ONE

Answer 1

Answers:

a) $1.05 rad/s$

b) $1.38 J$

Explanation:

a) Final angular velocity :

Before solving this part, we have to stay clear that the angular momentum $L$ is conserved, since we are dealing with circular motion, then:

$L_{o}=L_{f}$

Hence:

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

Where:

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

$\omega_{i}=0.73 rad/s$ 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 $I_{i}$ and $I_{f}$:

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

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

Where:

$I_{s}=8 kgm^{2}$ is the student's moment of inertia

$m=2 kg$ is the mass of each object

$r_{i}=1 m$ is the initial radius

$r_{f}=0.28 m$ is the final radius

Then:

$I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}$ (4)

$I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}$ (5)

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

$(12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}$ (6)

Finding $\omega_{f}$:

$\omega_{f}=1.05 rad/s$  (7) This is the final angular speed

b) Change in kinetic energy:

The rotational kinetic energy is defined as:

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

And the change in kinetic energy is:

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

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

$\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}$ (10)

Finally:

$\Delta K=1.38 J$ This is the change in kinetic energy