# Question on this problem

• lefthand
In summary, the conversation discusses using the formula 1/2mv^2+1/2lw^2=mgh to calculate the maximum height a mass will reach after a wheel is released and starts slowing down due to the downward tension of a cord. The formula is based on conservation of energy, but some possible mistakes to check for include using the wrong mass for calculating I, not accounting for the distribution of mass in the wheel, and using the wrong radius for calculating I or relating v and ω.

#### lefthand

Suppose the wheel is rotated at a constant rate so that the mass has an upward speed of 4.07 m/s when it reaches a point P. At that moment, the wheel is released to rotate on its own. It starts slowing down and eventually reverses its direction due to the downward tension of the cord. What is the maximum height, h, the mass will rise above the point P?
h =m

my question is what formula should i use? I've been using 1/2mv^2+1/2lw^2=mgh...but i haven't been getting the right answer...if anyone knows has a hint that would help i would like to use it :)

It appears that you are using the correct formula (conservation of energy), though without seeing the figure or your actual work I cannot be 100% sure. Here are a few things that you might check:

1. When you calculate I, do you use the mass of the wheel -- and not the rising mass?
2. Is the wheel like a solid disk, or is most of its mass in the rim? That will affect what I is, in terms of the wheel's mass and radius.
3. Sometimes a problem tells you the diameter of a wheel, and students can mistakenly use that instead of the radius to calculate I, or to relate v and ω to each other.

If you didn't make any of those mistakes I listed, then please post your work.