Rotational Equilibrium and dynamics

[ryomaechizen]

Please help me with these problems:

1) In a circus performance, a large 4.0 kg hoop with a radius of 2.0 m rolls without slipping. If the hoop is given an angeular speed of 6.0 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 15 degrees with the horizontal, how far (measured along the incline) does the hoop roll?

2) A wooden bucket filled with water has a mass of 75 kg and is attached to a rope that is wound around a cylinder with a radius of 0.075 m. A crank with a turning radius of .25 m is attached to the end of the cylinder. Wht minimum force directed perpendicularly to the crank handle is required to raise the bucket?

Thanks in advance.

 

[Doc Al]

Show your work and you'll get plenty of help.

Hints: For #1, use conservation of energy; for #2, consider the torques exerted on the cylinder.

 

[ryomaechizen]

For the first problem, we use the equation:
.5mv^2 + .5Iw^2 + mgh

right?

I set it up as (Mechanical energy initial = Mechanical energy final)
and I was wondering if this is the right way to set it up. For some reason, I can't get the right answer.

 

[Hootenanny]

I think you equation should be 1/2mv^2 + 1/2Iw = mgh beacuse all the initial linear kinetic energy and rotational kinetic energy will be converted into gravitational potential energy as the hoop travels up the incline.

 

[Doc Al]

For the first problem, we use the equation:
.5mv^2 + .5Iw^2 + mgh

right?

Right.

I set it up as (Mechanical energy initial = Mechanical energy final)
and I was wondering if this is the right way to set it up. For some reason, I can't get the right answer.

That's the right approach. Show what you did and we'll take a look. Two things to pay attention to: (1) how v relates to w, and (2) how distance up the incline relates to the change in height of the hoop's center of mass.

 

[Hootenanny]

I apologise for confusing the matter, but cud you please explain why the equation is +mgh and not =mgh

 

[Gamma]

I think he had written the total energy of the hoop.

Total Energy = Translational KE + Rotational KE + Grav. Potantial Energy.

= .5mv^2 + .5Iw^2 + mgh

which is the conserved quantity.

 

[Hootenanny]

Sorry, my mistake.