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?
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
To determine the speed of the cylinder after it falls 3.0m, we can use the principles of conservation of energy.
The potential energy of the cylinder at the beginning is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
When the cylinder falls to a height of 3.0m, its potential energy will be converted into kinetic energy. The formula for kinetic energy is KE = 1/2mv^2, where v is the velocity of the cylinder.
By setting the potential energy equal to the kinetic energy, we can find the velocity of the cylinder:
mgh = 1/2mv^2
The mass cancels out:
gh = 1/2v^2
Simplifying further:
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)
Now we can substitute the given values into the formula to find the velocity:
v = √(2 * 9.8 m/s^2 * 3.0 m)
v = √(58.8 m^2/s^2)
v ≈ 7.67 m/s
Therefore, the speed of the cylinder after falling 3.0m is approximately 7.67 m/s.