Question

A uniform solid disk has a mass of 1.00 kg and a radius of 1.00 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 7.00 rad/s. A uniform rod with a length of 3.00 m and a mass of 0.500 kg is released from rest, just above the turntable, such that the axis of the rod is the same as the axis of the disk. The rod slips on the turntable until it acquires the same final angular velocity.

a. Find the final angular velocity of the system.

b. Find the amount of mechanical energy lost due to friction.

Answer 1

Answer:

a) The final angular velocity of the system is 4 radians per second, b) The amount of mechanical energy lost due to friction is 5.25 joules.

Explanation:

a) The problem is a clear representation of the Principle of the Angular Momentum Conservation, where moment of inertia of the system is increased by the adding of the uniform rod and there are no external forces exerted on the system. This system is represented by the following model:

$I_{d} \cdot \omega_{o} = (I_{d} + I_{r})\cdot \omega_{f}$

Where:

$\omega_{o}$, $\omega_{f}$ - Initial and final angular velocities, measured in radians per second.

$I_{d}$, $I_{r}$ - Moments of inertia of the uniform solid disk and uniform rod, measured in $kg\cdot m^{2}$.

Now, the final angular speed is cleared:

$\omega_{f} = \frac{I_{d}}{I_{d}+I_{r}}\cdot \omega_{o}$

The moments of inertia of the uniform solid disk and uniform rod are modelled by these formulas:

Solid Disk

$I_{d} = \frac{1}{2}\cdot m_{d}\cdot r_{d}^{2}$

Where:

$m_{d}$ - Mass of the disk, measured in kilograms.

$r_{d}$ - Radius of the disk, measured in meters.

Given that $m_{d} = 1\,kg$ and $r_{d} = 1\,m$, the moment of inertia of the disk is:

$I_{d} = \frac{1}{2}\cdot (1\,kg)\cdot (1\,m)^{2}$

$I_{d} = 0.5\,kg\cdot m^{2}$

Rod (rotating about its center)

$I_{r} = \frac{1}{12}\cdot m_{r}\cdot l_{r}^{2}$

Where:

$m_{r}$ - Mass of the rod, measured in kilograms.

$l_{r}$ - Length of the rod, measured in meters.

Given that $m_{r} = 0.5\,kg$ and $l_{r} = 3\,m$, the moment of inertia of the rod is:

$I_{r} = \frac{1}{12}\cdot (0.5\,kg)\cdot (3\,m)^{2}$

$I_{r} = 0.375\,kg\cdot m^{2}$

Now, knowing that $\omega_{o} = 7\,\frac{rad}{s}$, the final angular velocity is:

$\omega_{f} = \left(\frac{0.5\,kg\cdot m^{2}}{0.5\,kg\cdot m^{2}+0.375\,kg\cdot m^{2}}\right)\cdot \left(7\,\frac{rad}{s} \right)$

$\omega_{f} = 4\,\frac{rad}{s}$

The final angular velocity of the system is 4 radians per second.

b) According to the Principle of Energy Conservation, the inclusion of the uniform rod on the turntable is represented by the following expression:

$K_{1} = K_{2} + \Delta E_{loss}$

Where:

$K_{1}$ - Rotational kinetic energy of the uniform disk, measured in joules.

$K_{2}$ - Rotational kinetic energy of the system (uniform disk + uniform rod), measured in joules.

$\Delta E_{loss}$ - Mechanical energy lost due to friction, measured in joules.

The mechanical energy lost due to friction is cleared:

$\Delta E_{loss} = K_{1} - K_{2}$

Now, the expression is expanded and mechanical energy losses is calculated:

$\Delta E_{loss} = \frac{1}{2}\cdot I_{d}\cdot \omega_{o}^{2} - \frac{1}{2}\cdot (I_{d}+I_{r})\cdot \omega_{f}^{2}$

$\Delta E_{loss} = \frac{1}{2}\cdot (0.5\,kg\cdot m^{2})\cdot \left(7\,\frac{rad}{s} \right)^{2} - \frac{1}{2}\cdot (0.5\,kg\cdot m^{2} + 0.375\,kg\cdot m^{2})\cdot \left(4\,\frac{rad}{s} \right)^{2}$

$\Delta E_{loss} = 5.25\,J$

The amount of mechanical energy lost due to friction is 5.25 joules.