Kinetic Energy of a Rotating Bar

In summary, the conversation discusses finding the kinetic energy of a thin and uniform bar rotating from one end on a pivot, with a length of 2.00m and weight of 12.0kg, that completes 5 rotations every 3 seconds. The conversation includes a hint to break the bar into infinitesimal segments and integrate to find the total kinetic energy. The conversation also mentions using the formula v=d/t=20∏m/3s and Ek=0.5*m*v^2 to calculate the speed and kinetic energy of each segment. The final solution involves integrating the function (0.5*v^2)dm, with dm expressed in terms of dx, to find the total kinetic energy of the bar.
  • #1
student34
639
21

Homework Statement



A thin and uniform bar has a length of 2.00 m and weighs 12.0 kg. It is rotating from one end on a pivot 5 times every three seconds. What is its kinetic energy?

This is its hint: break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energies of all of these segments.

Homework Equations



v = d/t = 20∏m/3s

Ek = 0.5*m*v^2

∫(12kg - 0kg)0.5*v^2dm


The Attempt at a Solution



∫(12kg - 0kg)0.5*v^2dm = (0.5(20∏/3)^3)/3 x 12 = 18374J

But I am way off because the answer is 877J
 
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  • #2
Have you seen the hint?
What is the speed of an element dm at distance r from the pivot?
What is its kinetic energy?
 
  • #3
nasu said:
Have you seen the hint?

No, the hint is just in words.

What is the speed of an element dm at distance r from the pivot?

V = (x*10*∏)/3

What is its kinetic energy?

0.5*(((x*10*∏)/3)^2)*dm

This seems like more advanced calculus than we are expected to know because we usually have to integrate functions that have only one variable, and it's usually obvious in the function. Or am I not seeing an easier way.
 
  • #4
No, you have only one variable, x.
You can write dm in terms of dx. Imagine the small mass element dm as a thin disk of thickness dx. The bar being uniform, of mass m and length L,
dm=(m/L)dx
 
  • #5
nasu said:
No, you have only one variable, x.
You can write dm in terms of dx. Imagine the small mass element dm as a thin disk of thickness dx. The bar being uniform, of mass m and length L,
dm=(m/L)dx

Oh great, I finally get it, thanks!
 

1. What is the definition of kinetic energy?

Kinetic energy is the energy an object possesses due to its motion.

2. How is kinetic energy calculated for a rotating bar?

The formula for calculating kinetic energy of a rotating bar is: KE = 1/2 * I * ω^2, where KE is kinetic energy, I is the moment of inertia of the bar, and ω is the angular velocity of the bar.

3. What factors affect the kinetic energy of a rotating bar?

The kinetic energy of a rotating bar is affected by its mass, moment of inertia, and angular velocity.

4. How does the kinetic energy of a rotating bar change as its angular velocity changes?

As the angular velocity of a rotating bar increases, its kinetic energy also increases. This is because the formula for kinetic energy includes the square of the angular velocity, so any changes to the angular velocity will have a greater impact on the kinetic energy.

5. Can the kinetic energy of a rotating bar be negative?

No, the kinetic energy of a rotating bar cannot be negative. It is always a positive value as it represents the energy of the object in motion.

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