b) At what angular speed does the bar/clay system rotate about its center of mas
ID: 1290110 • Letter: B
Question
b) At what angular speed does the bar/clay system rotate about its center of mass after the impact (in rad/s)?
On a frictionless table, a glob of clay of mass 0.420 kg strikes a bar of mass 1.680 kg perpendicularly at a point 0.190 m from the center of the bar and sticks to it.
a) If the bar is 1.500 m long and the clay is moving at 8.700 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 (in rad/s)?
On a frictionless table, a glob of clay of mass 0.420 kg strikes a bar of mass 1.680 kg perpendicularly at a point 0.190 m from the center of the bar and sticks to it. a) If the bar is 1.500 m long and the clay is moving at 8.700 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 (in rad/s)?Explanation / Answer
(a)
Vcm = sum(mi*vi) / total mass
in this case just one body has velocity:
Mclay = 0.42 kg
Mbar = 1.68 Kg
Mtotal = 2.1 kg
Vclay = 8.7 m/s
Vcm = 0.42 * 8.7 / 2.1 = 1.74 m/s
(b)
now you want to calculate the angular speed:
lets assume that you have a body who has a circular motion;
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 = 1.5 m
d = 0.19 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) + Mbar*(D/2) ]/Mtotal
Xcm = [0.42 *(0.19 + 0.75) + 1.68*(0.75) ]/2.1
Xcm = [0.3948 + 1.26) ]/2.1
Xcm = 0.788 m position of the center of mass
the angular velocit of the clay:
omegaclay = Vclay / D1
D1 = (d + D/2) - Xcm = (0.19 + 0.75) - 0.788
D1 = 0.152 m
omegaclay = 8.7/0.152 = 57.24 rad/s
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