Question

An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg m2 and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?

Please note: the professor has told us that the correct answer is 2.3 m/s. how does one arrive at this answer?

Answer 1

`v=2.28m/s`

Step-by-step explanation:

For the first mass we have
`m_1=3.5kg`, and for the second
`m_2=6.0kg`. The pulley has a moment of intertia
`I_p=0.0352kgm^2` and a radius
`r_p=0.125m`.

We solve this with conservation of energy. The initial and final states in this case, where no mechanical energy is lost, must comply that:

`K_i+U_i=K_f+U_f`

Where K is the kinetic energy and U the gravitational potential energy.

We can write this as:

`K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J`

Initially we depart from rest so
`K_i=0J`, while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:

`K_f=(m_1v^2)/(2)+(m_2v^2)/(2)+(I_p\omega_p^2)/(2)=(m_1+m_2+(I_p)/(r_p^2))(v^2)/(2)`

where we have used the rotational kinetic energy formula and that
`v=r\omega`

For the gravitational potential energy part we will have:

`U_f-U_i=m_1gh_(1f)+m_2gh_(2f)-(m_1gh_(1i)+m_2gh_(2i))=m_1g(h_(1f)-h_(1i))+m_2g(h_(2f)-h_(2i))`

We don't know the final and initial heights of the masses, but since the heavier,
`m_2`, will go down and the lighter,
`m_1`, up, both by the same magnitude h=1.25m (since they are connected) we know that
`h_(1f)-h_(1i)=h` and
`h_(2f)-h_(2i)=-h`, so we can write:

`U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)`

Putting all together we have:

`(K_f-K_i)+(U_f-U_i)=(m_1+m_2+(I_p)/(r_p^2))(v^2)/(2)+gh(m_1-m_2)=0J`

Which means:

`v=\sqrt{(2gh(m_2-m_1))/(m_1+m_2+(I_p)/(r_p^2))}=\sqrt{(2(9.8m/s^2)(1.25m)(6.0kg-3.5kg))/(3.5kg+6.0kg+(0.0352kgm^2)/((0.125m)^2))}=2.28m/s`