Answer:
[tex]v=2.28m/s[/tex]
Explanation:
For the first mass we have [tex]m_1=3.5kg[/tex], and for the second [tex]m_2=6.0kg[/tex]. The pulley has a moment of intertia [tex]I_p=0.0352kgm^2[/tex] and a radius [tex]r_p=0.125m[/tex].
We solve this with conservation of energy. The initial and final states in this case, where no mechanical energy is lost, must comply that:
[tex]K_i+U_i=K_f+U_f[/tex]
Where K is the kinetic energy and U the gravitational potential energy.
We can write this as:
[tex]K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J[/tex]
Initially we depart from rest so [tex]K_i=0J[/tex], 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:
[tex]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}[/tex]
where we have used the rotational kinetic energy formula and that [tex]v=r\omega[/tex]
For the gravitational potential energy part we will have:
[tex]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})[/tex]
We don't know the final and initial heights of the masses, but since the heavier, [tex]m_2[/tex], will go down and the lighter, [tex]m_1[/tex], up, both by the same magnitude h=1.25m (since they are connected) we know that [tex]h_{1f}-h_{1i}=h[/tex] and [tex]h_{2f}-h_{2i}=-h[/tex], so we can write:
[tex]U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)[/tex]
Putting all together we have:
[tex](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[/tex]
Which means:
[tex]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[/tex]
