On a frictionless table, a glob of clay of mass 0.30 kg strikes a bar of mass 0.
ID: 1907371 • Letter: O
Question
On a frictionless table, a glob of clay of mass 0.30 kg strikes a bar of mass 0.68 kg perpendicularly at a point 0.12 m from the center of the bar and sticks to it. A) If the bar is 0.58 m long and the clay is moving at 8.4 m/s before striking the bar, what is the final speed of the center of mass? B) At what angular speed does the bar/clay system rotate about its center of mass after the impact?
On a frictionless table, a glob of clay of mass 0.30 kg strikes a bar of mass 0.68 kg perpendicularly at a point 0.12 m from the center of the bar and sticks to it. A) If the bar is 0.58 m long and the clay is moving at 8.4 m/s before striking the bar, what is the final speed of the center of mass? B) At what angular speed does the bar/clay system rotate about its center of mass after the impact?Explanation / Answer
you will see that the velocity of the center of mass is:
Vcm = sum(mi*vi) / total mass
Vcm = 0.3 * 5.9 /(.3+.68) = 1.806 m/s
lets assume that you have a body who has a circular motion;
angular speed - 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 = 0.58 m
d = 0.12 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 = [.3 *(d+D/2)+.68*(D/2) ]/(.3+.68)
Xcm = 0.327 m position of the center of mass
the angular velocit of the clay:
omega clay = Vclay / D1
D1=(d+D/2)-Xcm = 0.083 m
omega clay = 8.4 / 0.083= 101.2 rad/s
now about the bar:
the center of the bar is situated related to the center of mass at:
Dbar = Xcm - D/2 = 0.037 m
If you assume that the bar when it rotate has the same tangential velocity v = 8.4 m/s
omega bar = 8.4 / 0.037 = 227.03 rad/s
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