# The two masses (m_1 = 5.0\ kg and m_2 = 3.0\ kg) in the Atwood's machine are released from rest,...

## Question:

The two masses ({eq}m_1 = 5.0\ kg {/eq} and {eq}m_2 = 3.0\ kg {/eq}) in the Atwood's machine are released from rest, with {eq}m_1 {/eq} at a height of {eq}0.91\ m {/eq} above the floor. When {eq}m_1 {/eq} hits the ground its speed is {eq}1.5\ m/s {/eq}. Assume that the pulley is a uniform disk with a radius of {eq}12\ cm {/eq}. Determine the mass of the pulley.

## Conservation of energy:

In an Atwood machine, with two masses, if the frictional forces are neglected then the reduction in the gravitational potential energy of the heavier mass is equal to the gain in the gravitational potential energy, kinetic energy and the rotational kinetic energy of the pulley.

Thus the total energy of the system is constant.

{eq}m_1 = 5\ kg\\ m_2 = 3\ kg\\ h = 0.91\ m\\ v= 1.5\ m/s\\ r = 0.12\ m\\ {/eq}

Let the mass of the pulley be M

When the heavier mass drops from rest then its gravitational potential energy is used up in moving the whole system. As a result the smaller mass also attains the height and the kinetic energy.

Thus, we can write, {eq}m_1\ g\ h = 0.5\ m_2\ v^2 + 0.5\ m_1\ v^2 + m_2\ g\ h + 0.5\times 0.5\ M\ r^2\ dfrac{V^2}{r^2}\\ {/eq}

Plugging in the given values, we get

{eq}5\times 9.81\times 0.91 = 0.5\times 3\times 1.5^2+ 0.5\times 5\times 1.5^2 + 3\times 9.81\times 0.91 + 0.25\times M\times 1.5^2\\ 44.6 = 3.375 + 5.625 + 26.78 + 0.5625\ M\\ \boxed{M = 15.68\ kg}\\ {/eq} 