One end of the string in the diagram at right is wrapped around the cylinder. The other end is tied to the ceiling. The cylinder is released from rest. How fast is it moving after it falls 3.0m?
2 answers
Please show the formula and how you do it. Thx
Look at energy:
potential energy lost= kinetic energy gained
mgh=1/2 m vf^2 +1/2 I w^2
mgh=1/2 m vf^2 + 1/2 I (vf/r)^2
Now, you have to decide the moment of inertia for the cylinder. Solid, or not. Lets just assume it is hollow for this demonstration.
I= m r^2
then
mgh= 1/2 m vf^2+1/2 m vf^2
Now, average velocity going down was Vf/2, so in three seconds it went 1.5Vf
h= 1.5Vf
1.5 mg Vf= 1/2 m Vf^2+ 1/2 m Vf^2
you can solve for Vf.
I would divide by mVf, and solve
potential energy lost= kinetic energy gained
mgh=1/2 m vf^2 +1/2 I w^2
mgh=1/2 m vf^2 + 1/2 I (vf/r)^2
Now, you have to decide the moment of inertia for the cylinder. Solid, or not. Lets just assume it is hollow for this demonstration.
I= m r^2
then
mgh= 1/2 m vf^2+1/2 m vf^2
Now, average velocity going down was Vf/2, so in three seconds it went 1.5Vf
h= 1.5Vf
1.5 mg Vf= 1/2 m Vf^2+ 1/2 m Vf^2
you can solve for Vf.
I would divide by mVf, and solve
2 answers