HELP Rotational kinetic energy problem

[Kudo Shinichi]

HELP!Rotational kinetic energy problem

Homework Statement

A 4.00kg mass and a 3.00kg mass are attached to opposite ends of a thin 42.0cm long horizontal rod. the system is rotating at angular speed omega=5.60rad/s about a vertical axle at the center of the rod. determine a) kinetic energy K of the system, and b) the net force on each mass. c)repeat parts a and b assuming that the axle passes through the center mass of the system.

The Attempt at a Solution

a)KE=1/2mv^2
m=4.00kg+3.00kg=7.00kg
v=radius*omega=21*5.60=117.6
therefore KE=48404.16J
b)Net force=F_t-F_f=F_t-mu*F_n
i don't really know how to get mu from this problem. and how do I get acceleration for F_t
c)total mass*radius for center mass=m1r1+m2r2=4*21+3*21=147
7*radius of center mass=147
therefore radius of center mass=147/7=21...
But it is same as question a and b
Can anybody help me with it. Thank you very much.

 

[berkeman]

b) If you have $$\omega$$, you should be able to calculate the centripital acceleration...

I think for c) and d), it just changes the radii that you use in your v^2 and centripital acceleration equations...

 

[Kudo Shinichi]

b) If you have $$\omega$$, you should be able to calculate the centripital acceleration...

I think for c) and d), it just changes the radii that you use in your v^2 and centripital acceleration equations...

centrpetal acceleration=omega^2*r=5.6^2*21=658.56m/s^2
Do I assume that there is no friction?
If yes then the net force should just be the force=ma
which are:
4*658.56=2634.24N
3*658.56=1975.74N

and for part c you said change the radii, but where should I change it to?the question states assume that the axle pass thru through the center mass of the system which I showed on my attempted question, but it still has the same length.

 

[berkeman]

The center of mass is not the center of the rod. Re-check your math. It's the same point on the rod where you could put your finger,and balance the rod+weights level. Think teeter-totter (the childs play thing where two kids sit on opposite ends of a board, with a fulcrum between them)...

 

[Kudo Shinichi]

The center of mass is not the center of the rod. Re-check your math. It's the same point on the rod where you could put your finger,and balance the rod+weights level. Think teeter-totter (the childs play thing where two kids sit on opposite ends of a board, with a fulcrum between them)...

Center of mass for24cm away from 3.00kg and 18cm away from 4kg object...but I didn't really know how to get it from a proper equation...I got the answer by plugging in the numbers and see whether the answers are eqaul to each other or not...sorry...I am not really good at physics.

 

[berkeman]

Center of mass for24cm away from 3.00kg and 18cm away from 4kg object...but I didn't really know how to get it from a proper equation...I got the answer by plugging in the numbers and see whether the answers are eqaul to each other or not...sorry...I am not really good at physics.

To balance the stick on your finger, you need the net torques from the two masses and lever arms to add to zero at your finger (the fulcrum). Write an equation for the sum of the torques at your finger, and you should start to see it. The equation you use for finding the center of mass of a linear object (like the stick and 2 weights) is basically the same equation.

Think about it. What's the equation for the torque about a point, in terms of the force (weight) and lever arm length? If you have two masses, one twice as heavy as the other, to balance them on a stick, the one weighing twice as much as the be _____ as far away from the fulcrum as the lighter one...