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 horse latitudes-high-pressure zones about 30 degrees north and 30 degrees south of the equator.-warm air traveling away from the equator cools and sinks-weather tends to be clear and drythe doldrums-low-pressure near equator-air rises and spreads out toward the poles-rising moist air produces clouds and heavy rainthe westerlies-blow from the west-move from the horse latitudes toward the poles.-bring storms across much of the united statesthe tradewinds-blow from the east-move from the horse latitudes toward the equator -strong, steady winds

Answer from: Quest

It's have to be 12 or idk

Another question on Physics

Physics, 21.06.2019 22:30

Astudent is given an assignment to demonstrate diffraction. he takes a photograph of a straw in a glass of water. the straw appears bent at the water level. which best describes this example? a) this is a good example of diffraction. b) this is an example of dispersion and not diffraction. c) this is an example of refraction and not diffraction. d) this is an example of reflection and not diffraction.

Answers: 1

Answer

Physics, 22.06.2019 16:50

Aresistor with an unknown resistance is connected in parallel to a 9 ω resistor. when both resistors are connected in parallel to an emf source of 21 v, the current through the unknown resistor is measured with an ammeter to be 4 a. what is the resistance of the unknown resistor? answer in units of ω.

Answers: 3

Answer

Physics, 22.06.2019 17:00

Abowling ball rolling down the lane toward the pins has gravitational potential energy. a. no b. a lot of c. a little

Answers: 2

Answer

Physics, 22.06.2019 21:00

During a car accident, a 125kg driver is moving at 31m/s and in 1.5s is brought to rest by an inflating air bag. what is the magnitude of the change in momentum to the driver

Answers: 2

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