Question

How long will it take for the heavier mass to reach the floor? What will be the speed of the two masses when the heavier mass hits the floor?

An Atwood’s machine has masses of 100 g and 110 g. The lighter mass is on the floor and the heavier mass is 75 cm above the floor.

Answer 1

`t approx 1.83s`

`v approx 0.82 ms^(-1)`

Explanation:

If we say that the heavier mass is `m_2` and the lighter is `m_1`, and we idealise so that, for instance the pulley has no inertia, then the magnitude of the acceleration of the masses is:

`a = g cdot (m_2 - m_1)/(m_2 + m_1) =g /22`

In terms of tile, the relevant equation of motion for the heavier mass (at constant acceleration) is:

`x = ut + 1/2 a t^2`

`t = sqrt((2x)/a)`

`t = sqrt((2 cdot 0.75)/(g/22)) approx 1.83s`

In terms of velocity, the relevant equation of motion for the heavier mass (at constant acceleration) is:

`v^2 = u^2 + 2 ax`

`v= sqrt(2 ax)`

`= sqrt(2 (g/22) 0.75) approx 0.82 ms^(-1)`