# Rotational Equilibrium and Dynamics

1. Feb 23, 2004

### Alethia

Ah jeeze, I am so sorry to be posting up three questions. I don't mean to just get answers for my homework, but would anybody be willing to lead me in the right direction regarding these physics problems? I have the solutions, but I am unsure on how to get them. Thank you, anything helps.

1) A solid sphere with a mass of 4.0 kg and a radius of 0.12 m starts from rest at the top of a ramp inclined at 15 degrees and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end. What is the total kinetic eneergy of the sphere when it reaches the bottom? (Assume that the sphere rolls without slipping and that g=9.81 m/s/s.)
--- When I first attempted this problem, I used KE(tot)=KE(rot) + KE(trans) to set things up, but I don't have the angular velocity, nor the acceleration/time.
--- The answer is 47 J

2) A 4.0 m board with a mass of 19 kg is pivoted at its center of gravity. A helium balloon attached 0.23 m from the left end of the board produces an upward force of 7.0 N. A 2.5 kg book is placed 0.73 m from the left end of the board, and another book of 1.3 kg is placed 0.75 m from the right end of the board. Find the torque on the board and the direction of rotation.
--- This problem is a very wordy one. I have my diagram drawn and everything. My first approach was to calculate torque for each side, seeing as it should be balanced (then net torque is zero). However, I ran into problems when calculating it because of the varying distances and masses (ie, what is the distance I use to calculate torque, do I configure in the board's mass, if so, how?)
--- The answer is 15 N*m

3) A force of 1250 N is needed to move a crate weighing 3270 N up a ramp that is 4.55 m long. If the elevated end of the ramp is 0.750 m high, what is the efficenicy of the ramp?
--- This problem seemed easy, but when I checked my answer, I was wrong. At frist I though I could just divide the to work output and work input, but I did not end up with the right answer, yet again. What confuses me the most, though, is when it says the crate weights #. How do I confiigure in the weight, along with the force required and distance?

2. Feb 23, 2004

### marcus

what's wrong with a straight application of PE = mgh

it comes down by h = 1.2 meters and that potential gets converted to some kind of kinetic energy (rotational or CM but they dont ask how much of each)

so at the end KE = mgh

(ps: it comes out right too. 4 x 9.81 x 1.2)

Last edited: Feb 23, 2004
3. Feb 23, 2004

### marcus

ideally with perfect efficiency it would take about 3270 x 0.75 joules of work to raise the weight up by 0.75 meter

but they say that using the ramp it takes 1250 x 4.55 joules of work
(the work it takes to push with 1250 N force all the way up the ramp which is 4.55 meters.)

compare the actual work 1250 x 4.55
with the work needed in a perfect world 3270 x 0.75

got to go, maybe someone will rephrase this nicer

4. Feb 23, 2004

### Alethia

Re: Re: Rotational Equilibrium and Dynamics

Thanks I get it! I never really thought of using potential energy. Silly me :p.

:D
Anybody for number two?

Last edited: Feb 23, 2004
5. Feb 24, 2004

### marcus

either I dont understand #2 right
or there is a typo
or an error in copying it

I assumed the center of gravity of the board
was at the midpoint----2 meters from either end.
And I got that the net torque was 2.8 newton meters counterclockwise
instead of 15. The answer you have there, 15 Nm, did not seem
right to me. better get someone else to take a stab at the
problem---I may be missing something

6. Feb 25, 2004

### turin

I concur.

7. Feb 25, 2004

### Alethia

Yeah whenever I solved it out, I would get 40 Nm, I think... or something. I'll ask the teacher about it. THanks!