- #1
MacLaddy
Gold Member
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Homework Statement
The mechanism in the attached photo contains two masses, each treated as particles. When released from rest, the 4-kg block falls and the 6-kg sphere rises. Initially, the angle between the two rigid, massless linkages is θ = 60°
What is the velocity of the sphere when θ = 180°?
Homework Equations
$$T_0 + V_0 = T + V$$[/B]
The Attempt at a Solution
I have two questions about my attempt at this problem. One is regarding my use of the datum line, and if I am using it correctly. The other is regarding the final velocity of the falling 4kg block. I am using the assumption that it will be zero, as it needs to change directions, but I am not sure if that is right. I have seen discussions on here in the past about the change in speed of a piston vs. the crank, but I don't recall what the final verdict was.
Anyhow, here is my attempt. Please let me know if I am on the correct track.
[itex]T_0 + V_0 = T + V[/itex]
Labeling the 4kg mass as "A", and the 6kg as "B", and using the pivot point just under the B mass as my datum line. Also assuming that [itex]T_0=0[/itex], as nothing is initially moving.
[itex]W_Ay_0 + W_By_0 = \frac{1}{2}m_Av_A^2 + \frac{1}{2}m_Bv_B^2 + W_Ay + W_By[/itex]
Assuming [itex]T_{A}=0[/itex]
[itex]W_Ay_0 + W_By_0 = \frac{1}{2}m_Bv_B^2 + W_Ay + W_By[/itex]
Solving for [itex]v_B[/itex] yields
[itex]v_B=\sqrt{2\frac{W_Ay_0 + W_By_0 - W_Ay - W_By}{m_B}}[/itex]
[itex]W_Ay_0=4kg(9.81 m/s^2)(-0.3m)=-11.772 kg*m^2/s^2[/itex]
[itex]W_Ay=4kg(9.81 m/s^2)(-0.6m)=-23.544 kg*m^2/s^2[/itex]
[itex]W_By_0=6kg(9.81 m/s^2)(0.3\sin(30)m)=8.829 kg*m^2/s^2[/itex]
[itex]W_By=6kg(9.81m/s^2)(0.3m)=17.658 kg*m^2/s^2[/itex]
Plugging everything in yields
$$v_B = 0.99 m/s \approx 1 m/s $$
Look about right?
Any advice is, as always, greatly appreciated.
Thanks,
Mac