Sphere On Incline (Rotational Kinematics and Energy)

  • Thread starter Pat2666
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Okay, another problem from me!

http://img187.imageshack.us/img187/69/32aa2d25kf9.jpg [Broken]

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 2.6 m down a q = 22° incline. The sphere has a mass M = 4.2 kg and a radius R = 0.28 m.


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a) Of the total kinetic energy of the sphere, what fraction is translational?
KEtran/KEtotal = *
0.71 OK


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b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?
KEtran = J *
28.66 OK


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c) What is the translational speed of the sphere as it reaches the bottom of the ramp?
v = m/s *
3.69 OK


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d) What is the magnitude of the frictional force on the sphere?
|f| = N

HELP: The frictional force provides the torque needed to give an angular acceleration. Therefore, first find the angular acceleration, then apply the rotational equivalent of Newton's 2nd Law (torque = I*a).
HELP: Since the sphere rolls without slipping, the angular acceleration is related to the translational acceleration, which can be found from kinematic relations.


Alrighty, so I managed to figre out all of the problem except this last bit. I thought I knew what to do after reading over the two HELP bits provided, but clearly I'm doing something wrong.

Since I know the Vt at the bottom of the ramp, and it being zero at the top of the ramp, I thought I could use that to solve for a (as you will see below). I then plug that into the modified equation for torque, but get the wrong answer. I don't fully understand how torque is equivilant to the force of friction, as that's what the HELP seems to say, but I'd also like to understand what I'm doing.

My Work :

http://img187.imageshack.us/img187/5945/workje4.jpg [Broken]
 
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Answers and Replies

  • #2
Doc Al
Mentor
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Since I know the Vt at the bottom of the ramp, and it being zero at the top of the ramp, I thought I could use that to solve for a (as you will see below).
Sure.
I then plug that into the modified equation for torque, but get the wrong answer.
What's the "modified" equation for torque? T = I*alpha. How does force relate to torque?
I don't fully understand how torque is equivilant to the force of friction, as that's what the HELP seems to say, but I'd also like to understand what I'm doing.
The friction is the only force that exerts a torque on the sphere (about its center of mass). But the force and the torque it produces are two different things. (They even have different units.)

Since you've found the linear acceleration, you can also just apply Newton's 2nd law for translation to find the friction force.
 
  • #3
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Ahhh, so I solved torque, but that's isn't to force, it's the product of the force and radius.

Thanks for the help :)

I knew something had to be wrong with my answer since I was pretty sure torque =/= the frictional force.
 

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