Kinetic Energy of a Rotating Bar

[student34]

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

 

[nasu]

Have you seen the hint?
What is the speed of an element dm at distance r from the pivot?
What is its kinetic energy?

 

[student34]

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.

 

[nasu]

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

 

[student34]

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!