# Homework Help: Torque and Motion Problem

1. Jun 11, 2010

### Hockeystar

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

All info is in attachment

2. Relevant equations

All info is in attachment

3. The attempt at a solution

I'm asking if someone can verify whether or not my work is correct. I'm doubting whether I solved the problem correctly or not.

#### Attached Files:

• ###### TorqueProblem.png
File size:
13.3 KB
Views:
119
2. Jun 11, 2010

### graphene

looks alright

3. Jun 11, 2010

### inky

I wonder table is smooth or not.

4. Jun 11, 2010

### hikaru1221

You got the wrong acceleration. Even if friction is neglected, the pulley has to be taken into account, as it's not massless. Here is my solution (assume that the table is frictionless):
_ For the 2-kg block:
$$m_2g - T_2 = m_2a$$
$$T_2 = 20 - 2a$$
_ For the 4-kg block:
$$T_4 = m_4a$$
$$T_4 = 4a$$
_ For the pulley:
$$T_2r - T_4r - M_f = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra$$
$$T_2 - T_4 - 2.5 = 1.5a$$
Thus:
$$20 - 2a - 4a - 2.5 = 1.5a$$
$$a = \frac{7}{3} m/s^2$$
_ The torque on the pulley:
$$M_p = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra$$
$$M_p = 0.7 Nm$$

Oh they ask for the torque on the pulley, which means this torque includes the friction's torque.

5. Jun 11, 2010

### inky

You mean tension on the string are not the same for 2 bodies. We consider same tension. According to hockeystar, you should take g= 9.8 ms-2.

6. Jun 11, 2010

### hikaru1221

It is impossible for the tensions to equal. Since the pulley is not massless, if the tensions equal, their torques on the pulley cancel each other, plus the frictional torque which prevents the pulley to rotate, the pulley will no way rotate.
If the pulley is massless, its moment of inertia is zero. From the equation: torque = (moment of inertia) x (angular acceleration), we shall see that torque = 0.

Sorry for not noticing g=9.8 m/s^2 :) So all to do is just recalculating.

7. Jun 11, 2010

### inky

Thanks . I got a lot of knowledge.