A Couple more "Rotation of Rigid Bodies" questions

In summary: So, 89rpm= 89/60= 1.483 rps. Then, centripetal acceleration at the rim is given by a=v^2/r= (1.483)^2(3.63)= 8.65 m/s^2. For the second part, the initial potential energy is mgR while the final kinetic energy is (1/2)Iω^2= (1/2)(1/2)mR^2(ω^2). Since the disk is rotating about an axis through its center, the moment of inertia of the disk is (1/2)mR^2. So, setting these two equal, we get mgR= (1/4)mR^
  • #1
sophixm
3
0

Homework Statement


1.
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m3 , in the shape of a uniform disk with a thickness of 11.5cm . (also KE=10.6MJ, angular velocity=89rpm)
What would be the centripetal acceleration of a point on its rim when spinning at this rate?
2. A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass m is glued to the rim of the disk.
If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.
Express your answer in terms of the variables m, R, and appropriate constants

Homework Equations


1. Centripetal acceleration= v^2/r
v=rω
2. Conservation of Energy
U=mgR
K=1/2Iω^2[/B]

The Attempt at a Solution


1. so in part one I found the diameter to be 7.26m through sheer luck. I'm going into office hours this week to understand this part of the problem, but I won't have time to before the assignment is due. The second part I don't have seems easy enough, I converted 89.0rpm to rps by dividing by 60, so 1.48rps, times the radius 3.63, square the answer, and divided by the radius. I know my answer is wrong though because the answer to the problem in the book is 327m/s^2 with slightly different values, and i get a very small value.
2. I've done some searching online, and I'm pretty sure I'm just supposed to set the initial potential energy equal to the final kinetic energy, so mgR=(1/2)Iω^2. I think I'm just simplifying wrong, but with my incorrect answers I've gotten the feedback that the correct answer does not depend on: Rm, m, or mR. Using different solutions if come up with different sources, I've tried √(19.6m/Rm), √(2g/3R), and √(9.8m/.5mR), and obviously none of those are right.
Thanks for any help:)
 
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  • #2
sophixm said:

Homework Statement


1.
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m3 , in the shape of a uniform disk with a thickness of 11.5cm . (also KE=10.6MJ, angular velocity=89rpm)
What would be the centripetal acceleration of a point on its rim when spinning at this rate?
2. A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass m is glued to the rim of the disk.
If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.
Express your answer in terms of the variables m, R, and appropriate constants

Homework Equations


1. Centripetal acceleration= v^2/r
v=rω
2. Conservation of Energy
U=mgR
K=1/2Iω^2[/B]

The Attempt at a Solution


1. so in part one I found the diameter to be 7.26m through sheer luck. I'm going into office hours this week to understand this part of the problem, but I won't have time to before the assignment is due. The second part I don't have seems easy enough, I converted 89.0rpm to rps by dividing by 60, so 1.48rps, times the radius 3.63, square the answer, and divided by the radius. I know my answer is wrong though because the answer to the problem in the book is 327m/s^2 with slightly different values, and i get a very small value.
2. I've done some searching online, and I'm pretty sure I'm just supposed to set the initial potential energy equal to the final kinetic energy, so mgR=(1/2)Iω^2. I think I'm just simplifying wrong, but with my incorrect answers I've gotten the feedback that the correct answer does not depend on: Rm, m, or mR. Using different solutions if come up with different sources, I've tried √(19.6m/Rm), √(2g/3R), and √(9.8m/.5mR), and obviously none of those are right.
Thanks for any help:)
Angular velocity is usually expressed in units of radians per second, rather than revolutions per second .
 

1. What is meant by "rotation of rigid bodies" in physics?

The rotation of rigid bodies refers to the movement of an object in which all of its particles move in circular paths around a fixed axis. This type of motion is characterized by the fact that the distance between any two particles of the object remains constant throughout the rotation.

2. How is angular velocity related to the rotation of rigid bodies?

Angular velocity is a measure of the rate at which an object rotates around an axis. It is directly related to the rotation of rigid bodies as it determines the speed at which each particle of the object is rotating around the axis.

3. Can a rigid body rotate around more than one axis at a time?

Yes, a rigid body can rotate around multiple axes simultaneously. This is known as compound rotation and can be seen in objects such as a spinning top or a gyroscope.

4. How is rotational kinetic energy calculated for a rigid body?

The rotational kinetic energy of a rigid body is calculated using the formula 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity of the object. The moment of inertia is a measure of the object's resistance to rotational motion.

5. What is the difference between rotational and translational motion?

Rotational motion refers to the movement of an object around an axis, while translational motion refers to the movement of an object in a straight line. In rotational motion, the distance between any two points of the object remains constant, while in translational motion, the object's position changes over time.

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