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

Answer:

$v=2.28m/s$

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:

$K_f=\frac{m_1v^2}{2}+\frac{m_2v^2}{2}+\frac{I_p\omega_p^2}{2}=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}$

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:

$(K_f-K_i)+(U_f-U_i)=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}+gh(m_1-m_2)=0J$

Which means:

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