Smartguy94
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Homework Statement
On a frictionless table, a glob of clay of mass 0.72 kg strikes a bar of mass 1.34 kg perpendicularly at a point 0.23 m from the center of the bar and sticks to it.
If the bar is 1.22 m long and the clay is moving at 8.3 m/s before striking the bar, what is the final speed of the center of mass?
At what angular speed does the bar/clay system rotate about its center of mass after the impact?
2. The attempt at a solution
Here is what I did
Vcm = sum(mi*vi) / total mass
in this case just one body has velocity:
Mclay = .72 kg
Mbr = 1.34 Kg
Mtotal = 2.06 kg
Vclay = 8.3 m/s
Vcm = 2.901 m/s
which is correct, I got the first part, but here is the second part that I got wrong
now calculate the angular speed:
angular speed - omega
omega = v /r
v - is the linear velocity on the trajectory of the body (tangential velocity)
r - is the distance between the body which rotate and the center of rotation
in your case the centre of rotation is the center of mass and r is the distance of the clay to the centre of mass
D = 1.22 m
d = 0.23 m
the centre of the bar related to one end is D/2
the position of the clay related to the same end is d+D/2
Xcm = [Mclay *(d+D/2)+Mbr*(D/2) ]/Mtotal
Xcm = .6904 m position of the center of mass
the angular velocit of the clay:
omega clay = Vclay / D1
D1=(d+D/2)-Xcm = .1496 m
omega clay = 55.477 rad/s
now about the bar:
the center of the bar is situated related to the center of mass at:
Dbar = Xcm - D/2 = .0804 m
If you assume that the bar when it rotate has the same tangential velocity v = 8.3 m/s
omega bar = 8.3 / 0.0804 = 103.249 rad/s
the question is asking about At what angular speed does the bar/clay system rotate about its center of mass after the impact?
and so I add up both the omega of the bar and the clay and got 158.726 rad/s
but it's wrong
Can anyone tell me where my mistake is?