Rotation of a hoop about an off-center axis

[Chad Jensen]

Homework Statement

I am studying for a test and am struggling with this question on the practice exam. The answer is 9.9 I am not sure how she got to the answer though
A hoop of mass 300 g. and a radius of 20 cm rotates about an axle at the edge of a hoop. The hoop starts at highest position(pi/2) and is given a very small push. What is the angular speed when the hoop is in the lowest position(3pi/2)

Homework Equations

[/B]
I am not sure about the relevant equations. I am sure it will take the formula for I=mr² and I am pretty sure it takes the off-center axis of rotation. I_new=I_{center mass} +Md² I am not sure the common thread they have to tie them together. Is torque also involved? If it is it would have to be an integral because the pull of gravity on the hoop varies with sin(θ) and we haven't integrated except on one exam this year, so i doubt that's the route.

The Attempt at a Solution

I knew it wasn't right but I tried the formula W²_f=W²_i+2αΔθ.

Edit: Here are my thoughts after looking at it again. I can find "I" and there are 3 equations given on my exam formula sheet that have "I" as a factor; Στ=Iα for torque, L=Iω for momentum and 1/2Iω² for energy. I believe you tie everything together with the energy formula. Is that right?

Any help would be appreciated.
Thank you
Chad

 

[Mitchel Haas]

You'll want to use conservation of energy to solve this, as it's the easiest approach.
U1 + K1 + W = U2 + K2.

The initial energy is just that of the height of the center of mass (2R), m g h = (.3)(9.8)(2*.2) = 1.176J.
The ending energy is the rotational KE of the hoop about the axis, or (.5) I ω².
To calculate I, note that the CM is not the center of the hoop, since the axis is at the rim of the hoop, so you need to use the Parallel-Axis Theorem Ip = I_cm + Md².
For a hoop, this would be I = MR² + Md² = MR² + MR² = 2MR².
So, we have 1.176J = .5 * 2MR²ω² = (.5)(2)(.3)(.2^2)ω² = .012ω².
Solving for angular speed squared, ω², ω² = 1.176/.012 = 98.
The square root of 98 is 9.899, or 9.9.

You can also solve this with the ending energy being the 'translational' KE of the CM of the hoop, plus the rotational KE of the hoop about the CM of the hoop.
This would be a little more complex, but you'd end up with the same answer.

 

[BvU]

Hello Chad,

A bit of advice from an old hand, somewhat improving on Mitchel ( back, I may say ):

Work with symbols as long as you can; do not start calculating unless it is necessary somehow. Check dimensions before calculating.
In this case you begin with an energy balance $${1\over 2}I\omega^2 = m\;g\;\Delta h = m\;g\; 2r $$ and you have ##I = mr^2 + mr^2## so you are left with $$\omega^2 ={ m\;g\; 2r \over mr^2 } = {2 \;g\over r} $$ which has the dimension of seconds^-2, so that's OK. And less risk of typing mistakes or wrong rounding off if you calculate now.

 

[Chad Jensen]

Thank you for the excellent answers. I really appreciate it. I have a hard time right now with using variables and setting up the correct formula. It is easier right now to see numbers because I know where they need to go most of the time. With variables, I have a hard time knowing what goes where. It will come with practice I am sure but it is so stressful until I get the skill developed