Suppose three masses are arranged as shown, connected by a rodless mass.
(2 kg)-------(6 kg)--------(4 kg)
2m 3m
a) If this object is free to rotate in space, about what point will it spin?
b) A small mass (m=0.1 kg) drops vertically onto the2 kg mass traveling 100 m/s and buries intself into the mass. What is the period of rotation of the system immediately after impact?
c) What is the rotational kinectic energy of the system after the impact?
2 answers
I'm sorry, 2m is the distance betweek 2 kg and 6 kg. 3m is the distance between 6kg and 4kg!
a. it spins about its center of mass. Where is that? From the left end,
cm=(2*0+6*2+4*5)/12kg=28/12 m from the left end.
b. conservation of angular momentum:
momentum before=momentum after
Io (v/r)=Itotal*w
.1*r^2*100/r=Itotal w
now what is r? well, r is the distance from the left end to the cm, or 28/12 m
What is I total?
it is the sum of the three masses.
I total= Ileft + I big + I right
= (2.1kg)(28/12)^2+ 6kg*(28/12-2)^2+ 4kg*(5-28/12)^2
solve for w, in radians per second.
Period= 2PI/w
cm=(2*0+6*2+4*5)/12kg=28/12 m from the left end.
b. conservation of angular momentum:
momentum before=momentum after
Io (v/r)=Itotal*w
.1*r^2*100/r=Itotal w
now what is r? well, r is the distance from the left end to the cm, or 28/12 m
What is I total?
it is the sum of the three masses.
I total= Ileft + I big + I right
= (2.1kg)(28/12)^2+ 6kg*(28/12-2)^2+ 4kg*(5-28/12)^2
solve for w, in radians per second.
Period= 2PI/w
1 answer