Sunday
March 26, 2017

Post a New Question

Posted by on .

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?

  • Physics 121 - ,

    Please show the formula and how you do it. Thx

  • Physics 121 - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question