## IOLab Atwood’s Machine

### Angle of slope

The spring is put on the IOLabs’s force probe. Tilt a book and shim one end and put it’s box at the end of the book so that the force probe rests on the box and measures the force while it’s at rest on the end block.

$$F_{probe} = mg\sin\theta$$

$$\theta = \arcsin{\frac{F_{probe}}{mg}}$$

(You’ll note that we’re not being completely consistent with signs here, but that’s ok if we know what we’re doing. 🙂

In class we measured $F_{probe} = 0.18 $ N. The mass of the cart is about 200 g= 0.200 kg. Therefore θ = 0.092 radians = 5.3°.

## Average force while bouncing

If you start the cart rolling down the slop a distance $x$ up the ramp it will take $\Delta t$ time to reach the box and bounce back.

While accelerating down the ramp the acceleration is $a = g \sin\theta$. Therefore the velocity when the cart hits the box is

$$v_i = (g\sin\theta)\Delta t$$

For simplicity assume that there is no kinetic energy lost during the bounce and the speed just after the bounce is the same as just before, but in the opposite direction.

$$v_f = -(g\sin\theta)\Delta t$$

Therefore, the momentum change during the bounce is

$$\Delta p = m(v_f – v_i)$$

$$\Delta p = (2\Delta t) mg\sin\theta$$

The force bumps while bouncing occur at an interval of $2\Delta t$ because it take $\Delta t$ to fall and another $\Delta t$ to go back up if we no energy is lost in the bounce. So the change in momentum must equal the average force. :

$$\int_{t_1}^{t_2} F(t) dt = F_ {\rm avg} (t_2 – t_1)$$

Take $t_1$ and $t_2$ be at the peaks of consecutive force bumps: $(t_2-t_1) = 2\Delta t$.

$$\Delta p = (2\Delta t) mg\sin\theta =F_ {\rm avg} (2\Delta t)$$

Therefore

$$F_ {\rm avg} = mg\sin\theta$$.d

The effects of energy loss during the collision and friction while the cart is rolling up and down are being neglected. However, the experimental result that the average force while bouncing is about the same as the resting force suggests that when these effects are taken into account, we’ll get the same answer.

## Atwood’s Machine with the IOLab

We’re going to grossly idealise the situation by neglecting friction. (We could have used the superpully, but this was easier to do quickly.)

Since the two masses are connected with strings, the accelerations are the same.

Neglecting the friction we assume the $F_T$, the tension in the string is the same on both ends.

(Friction of the string on the table will negate this constraint.)

Using the wheel we got $a=0.87$ m/s^{2}. Therefore the tension had to be

$F_T = m_2 a$ = 0.174 N. The actual measured force was 0.20 N. So it’s reasonable that the wheel friction could be 0.026 N. ($\mu_k \approx 0.0013$)