Physics, 05.07.2020 04:01 brendonvernon8

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.

Answers: 3

Show answers

Answers

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.

Answer from: Quest

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

Answer from: Quest

with the invention of the automobile, the woman of the house could drive to the store and get groceries. women were also free to socialize more. during world war ii women were allowed to work in factories to the war effort and could drive to work or ride mass transit buses to their jobs.

Another question on Physics

Physics, 21.06.2019 23:30

Ablock of mass m = 2.5 kg is attached to a spring with spring constant k = 790 n/m. it is initially at rest on an inclined plane that is at an angle of 28 degrees with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is k = 0.14. in the initial position, where the spring is compressed by a distance of a = 0.18 m, the mass is at its lowest position and the spring is compressed the maximum amount. take the initial gravitational energy of the block as zero. b) if the spring pushes the block up the incline, what distance, l, in meters will the block travel before coming to rest? the spring remains attached to both the block and the fixed wall throughout its motion.

Answers: 1

Answer

Physics, 22.06.2019 16:30

Humidity is to blame for that muggy, steamy feeling you experience on some hot summer days. what gas in the atmosphere causes humidity? a) oxygen b) hydrogen c) nitrogen d) water vapor

Answers: 1

Answer

Physics, 23.06.2019 05:00

Definition: this is the weighted average of all of the isotopes of an element.

Answers: 1

Answer

Physics, 23.06.2019 20:00

Answer each myth with facts and evidence, explain why they are un-true. 4. myth: an organism's kingdom only describes physical characteristics. fact: evidence: 5. myth: mammals and plants don't belong in the same domain. fact: evidence:

Answers: 1

Answer

You know the right answer?

A uniform solid disk has a mass of 1.00 kg and a radius of 1.00 m. The disk is mounted on frictionle...

Questions