# Forces, Torques, and Acceleration

1. Feb 16, 2004

### Arfrce8729

Hi I have the following homework problem that I need to solve two times, one time using forces / torques / acceleration and again using work and energy. I have attached a diagram of the problem and the variables. Pertaining to the first method, I think the solution could be achieved by the following...

( The friction of the pulley is neglected, but the connecting strin does not slip )
( Angular acceleration will be represented by a<ANG> )

Starting from Newton's Second law,

F = m * a

The equations for the block on the table and the hanging block are as follows...
(T is the tension of the string)

<table block>

T - fk = m<table> * a

<hanging block>

T - m<hanging> * g = m<hanging> * a

The torque of the pulley is:
(I is the rotational inertia)

Torque<NET> = I * a<ANG>

which equates to:

-R * T = 1/2 * M<pulley>(r^2 + R^2) * a<ANG>

(R is negative because it's moving in a clockwise direction. I is rotational intertia of an annular cylinder)

To eliminate T, the net force equation for the table block is solved for T:

T = m<table> * a + coeff-kin * m<table> * g

I'm not really sure which direction to go after this, or if this is even correct. I think I need to substitute the above equation for T into the annular cylinder equation, but that doesn't look like it works out corectly. Anyone have any ideas? Thanks in advance for any help.

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2. Feb 17, 2004

### jamesrc

Since we're considering the dynamics of the pulley, the tension on either side of the pulley is not equal. You've got some things right, but I'm just going to start at the beginning:

Let's call the acceleration of the blocks a and say that a positive value for a involves the block on the table (m1) moving to the right. This means that a positive a also involves the hanging block (m2) moving down. We don't consider slip of the cable, so this means that the pulley must have an angular acceleration of &alpha; = a/R, where R is the outer diamater of the annular disk (pulley). Note that for positive a, &alpha; is counterclockwise. Keep these conventions in mind as we go through the free body diagrams:

From a FBD for m1:

$$T_1 - \mu m_1 g = m_1 a$$

From a FBD for m2:

$$m_2g - T_2 = m_2a$$

From a FBD of the pulley:

$$(T_2 - T_1)R = I\frac a r = \frac{M_p(R^2+r^2)}{2}\frac{a}{r}$$

I hope it is clear where these equations came from (if not, please ask). Mp is the mass of the pulley, and R and r are the outer and inner radii of the pulley, respectively.

There, you have 3 equations and three unknowns. Once you solve for a, you can use basic kinematics for constant acceleration motion to find the speed at the position given.

Post back if you need help with this or with the energy method solution.

Last edited: Feb 17, 2004
3. Feb 17, 2004

### Arfrce8729

Ah thanks a bunch. That makes everything a lot clearer. I think that I might need a little help for the solution using work and energy. Since friction, a nonconservative force, is present I can't use the conservation of mechanical energy in this problem right? Instead I would have to use

Work = Change in Total Energy = Change in Mechanical Energy + Change in Thermal Energy

possibly? But how does the pulley fit into this? Thanks...

4. Feb 17, 2004

### jamesrc

You could use the work energy theorem. The work done on the system is equal to the change in total energy. Let's use the subscripts 1 and 2 to indicate the initial and final states of the system. We can also set the datum to be the initial height of m2.

So what's E1, the total initial energy of the system? Nothing is moving, so there's no kinetic energy. According to the datum we set, m1 does have some gravitational potential energy, but we know that that mass does not change height, so we can ignore that in this problem.

After the system has moved a distance d, all 3 moving objects have kinetic energy. The total kinetic energy is given by:

$$K_2 = \frac{m_1v_1^2}{2} + \frac{m_2v_2^2}{2} + \frac{I\omega^2}{2}$$

(I is the inertia of the pulley) We know from the geometry of the system that v1 = v2 = &omega;R = v, the velocity we're solving for.

The final gravitational potential energy is -m2gd

So the work-energy theorem tells us that:

$$\frac{m_1v^2}{2} + \frac{m_2v^2}{2} + \frac{M_p(R^2+r^2)v^2}{4R^2} - m_2gd = -\mu m_1gd$$

The right hand side of the equation is the work done by friction on the system.