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dawidjoubert

Angular Energy?

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How can i find the amount of energy in a spinning sphere(of equally distributed mass)? If i know that it does say 2pi a second (1 revolution a second) how can i find the amount of kinetic energy that is. Could sum1 please help me. Mass = 1kg Angular Velocity = 5(2 * PI)/s (5 revolutions a second) What is the kinetic energy? (or angular energy, i just need a number in newtons) If needed i can find the radius

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Angular kinetic energy is defined to be (1/2)Iw^2.
I is the inertia. For a sphere rotating about a central axis it is (2/5)Mr^2.
r is the radius of the sphere.
w is the angular velocity.

All together you get (1/5)(Mr^2)w^2.

Newtons also measure linear force, not angular energy. Energy is in Joules (also known as Newton-Meters, or kg*m^2/s^2)

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I understand newtons is linear force, but you see i need to prove the original potential energy is equal to the new kinetic & angular energy (and ofcourse i will subtract friction)

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Here is what I am doing:
http://rmcommando.co.uk/images/tempPhysics.jpg



Radius = 1cm = 0.01m
Mass = 1kg

A = gSin23.35 = 3.884m/ss
T = (2d/a)^1/2 = 0.7564
V = 2d/t = 2.934m/s
Kinetic Energy = 1/2mv^2 = 4.304 joules
Potential Gravity = 0.44m * g * m =4.312


Angular Properties
U = 0 T = 0.7564 V = (2.2m / 0.01m)/t = 293 Revs

Angular Velocity = 2pi * V = 1840 radians/s
Angular Kinetic Energy = 1/5Mr^2W^2=67.712 joules

How can the angular kinetic energy be so many joules?

[Edited by - dawidjoubert on March 11, 2006 5:59:04 PM]

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Quote:
Original post by dawidjoubert
A = gSin23.35 = 3.884m/ss

Your troubles start right there, where you are treating the sphere as a particle. This is what the acceleration would be if the sphere slid down a frictionless plane without rolling.
To solve this, I suggest you look at the energies before and after. Come up with an expession for the energy of the rolling ball (hint: integrate the kinetic energy of every small element in the sphere).
In the end, you should get something like this
m*g*h = 0.5*m*(R*w)^2 + 0.5*Ig*w^2
w - angular velocity, radians per second
Ig - moment of inertia. As mentioned, for a sphere this is (2/5)Mr^2

Where you can solve for w

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Quote:
Original post by Ksero
Quote:
Original post by dawidjoubert
A = gSin23.35 = 3.884m/ss

Your troubles start right there, where you are treating the sphere as a particle. This is what the acceleration would be if the sphere slid down a frictionless plane without rolling.
To solve this, I suggest you look at the energies before and after. Come up with an expession for the energy of the rolling ball (hint: integrate the kinetic energy of every small element in the sphere).
In the end, you should get something like this
m*g*h = 0.5*m*(R*w)^2 + 0.5*Ig*w^2
w - angular velocity, radians per second
Ig - moment of inertia. As mentioned, for a sphere this is (2/5)Mr^2

Where you can solve for w


Um okay, the ball starts at rest. So it doesn't start rotated.
Let me quickly try that!
So that is

4.312N = 1/2(0.0001 * w^2) + 1/5(0.0001*w^2)
Factorise
4.312N = w^2(1/20000 + 1/50000)

w = sqrt(4.312/(1/20000+1/50000) = 248radians/sec

So thus the Angular Energy is:
Angular Kinetic Energy = 1/5Mr^2W^2=1.232 Joules

Now that sounds more realistic! Is that correct?

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