# Homework Help: Rotation with friction

1. Apr 1, 2004

### cristina

A 4kg block resting on a horizontal ledge is attached to his right to a string that passes over a pulley and is attached to a hanging 2kg block. The coefficient of friction between the ledge and the 4kg block is 0.25. The pulley is a uniform disk of radius 8cm and mass 0.6kg. a) Find the speed of the 2kg block after it falls from the rest a distance of 2.5m. b) What is the angular velocity of the pulley at this time?

Delta K + Delta U = 0
Because Ki=Uf = 0
½(m+M)v^2+1/2(I,pulley)w^2-mgh=0
½(m+M)v^2+1/2(1/2MR^2)(v^2/R^2)-mgh=0

so v = sqroot(2mgh/(M+m+1/2M,p))
v = sqroot(2(2kg)(9.81)(2.5)/(4kg+2kg+1/2(0.6kg))) =3.95m/s

b) w= v/r = (3.95m/s)/0.08m) = 49.5rad/s

I did solve it without the 0.25 coefficient of fiction though may you help me please?

Last edited: Apr 2, 2004
2. Apr 1, 2004

### NateTG

How much work does the friction do?

3. Apr 1, 2004

4. Apr 1, 2004

### NateTG

No.

Work = Force X Distance

It will be an extra term in the energy equation.

5. Apr 1, 2004

### NateTG

The work done by friction is going to be force times distance.

This would be easier to illustrate with a drawing, but it's going to be:
$$F_f=mg\mu_k$$
so
$$F_f=4 (9.81) 0.25=9.81$$
mow
$$W_f=9.81*2.5$$
you can plug that into your energy equation:
$$\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}I\omega^2+W_f-mgh=0$$

I think you've got the math under control from there.

6. Apr 1, 2004

### cristina

thank you very much, I got it.

7. Apr 2, 2004

### NateTG

$$E_{final}=E_{inital}$$
The work done by friction typically ends up as a temeperature change on the $$E_{final}$$ side. It's just one more thing that you need to account for in the energy equation