# Thin rod

1. Nov 29, 2009

### seraphimhouse

1. The problem statement, all variables and given/known data

Attached to each end of a thin steel rod of length 1.00 m and mass 6.80 kg is a small ball of mass 1.15 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 43.0 rev/s. Because of friction, it slows to a stop in 25.0 s. Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the 25.0 s.

2. Relevant equations

K = Kf - Ki = (1/2)Iwf^2 - (1/2)IWi^2 = W
T = Ia (a = angular acceleration)

3. The attempt at a solution

I found the angular acceleration to be -10.816 rad / s^2
and the number of revolutions to be 537.42 rev

For torque i used the equation T=Ia where I = 1/12ML^2 where M is the total mass of the balls attached to each end of the rod plus the mass of the rod itself which is a total of 9.10 kg. I got T = -8.202 [N m] and for the work I used the work equation but got a really high solution which was -27723 J.

Both are wrong according to my homework's website

2. Nov 29, 2009

### kuruman

You can't just add the masses of the balls to the mass of the rod the way you did. The moment of inertia of the system is the moment of inertia of the rod (1/12)ML2 to which you add the moments of inertia of the two masses m(L/2)2+m(L/2)2.

3. Nov 29, 2009

### seraphimhouse

so would m be half the mass of the rod plus the mass of one ball?

4. Nov 29, 2009

### kuruman

No. M in my expression is the mass of the rod (6.80 kg) and m is the mass of one ball (1.15 kg). Then

$$I=\frac{1}{12}ML^2+2 \times m(\frac{L}{2})^2$$

where L = 1.0 m.

5. Nov 29, 2009

Thank you!