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Rolling solid sphere question

  • Thread starter mighty2000
  • Start date
39
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Hey there.

My question is:

A solid sphere of mass 2.5 kg and radius R =0.20 m is initially rolling along the horizontal without slipping with a center of mass speed V=6.0 m/s

1) if the sphere then rolls without slipping up a fixed ramp(25 deg) and a coefficient of static friction of .35, how far X(in meters) up the ramp does it roll?

Anyway I think I figured this one out correctly, but Im not too certain what to do with the friction. Do I just add it in to the final answer or what?

my work:

Ei=1/2 Mv^2 + 1/2 I(omega)^2
Ef=mgh
Ef=Ei

becomes 1/2 v^2 + 1/5 v^2 = gh (cancel all mass)

becomes h=(v^2/2g)+(v^2/5g)

end up with a height of 2.56 meters up the ramp

2)
Now I am to find the sphere's center of mass acceleration up the ramp

I used a(cm)=-a(t)=-r(alpha)
fs=(I*a(cm))/r^2
a(cm) = gsin(theta)-I*(a(cm)/mr^2)
a(cm) = gsin(theta)/(1+(I/mr^2))

a(cm)=2.961 m/s^2

3)
and finally I am to calculate the static friction acting on the sphere as it rolls up the ramp

I used the same formula as above and plugged in the number a(cm) and of course I got the same answer of 2.961 N

Any help on this problem would be appreciated.

Thanks in advance

M2k
 
512
0
Static friction? AFAIK, you only need to calculate static friction when you want to know if an object starts to slip or not. Here, it says 'without slipping'. I think static friction plays no role here, so your 1)st answer is correct.
 
39
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what about the rest bro?
 
512
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I think you don't need the rest. The question is, how far does it roll, and you answered that.
 
39
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No dude, the other two are also questions that I have to answer.
And from what you are saying, the answer to number 3 is zero, because it is not slipping. Right?

Thanks for the reply
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
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Originally posted by arcnets
Static friction? AFAIK, you only need to calculate static friction when you want to know if an object starts to slip or not. Here, it says 'without slipping'. I think static friction plays no role here, so your 1)st answer is correct.
No, static friction is what moves a rolling object forwards. It plays the most important role of all!

M2k:

To answer the last two parts, draw a free body diagram of the ball on the ramp. Then write down Newton's second law, which is not 2 but 3 equations (you sum the forces and the torques). Don't forget that you have static friction acting in the direction of motion of the center of mass of the sphere.

Try to write down and solve the system of equations. You should be able to get 2 and 3 simultaneously.
 
39
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so f(s) is actually moving with the sphere?

once again Tom, do I just add the coefficent of sf on to the answer or do I have to say:

a(y)=0 mu(s)N = mu(s)*mgsin(theta) and that is an additional force since it is moving with the sphere. I understand that the friction is what is required for the sphere to not slip.

Sorry bro, more help required, I drew out the free body diagram.

Thanks for the reply

M2k
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,475
20
Originally posted by mighty2000
so f(s) is actually moving with the sphere?
The frictional force always acts at the point of contact of the sphere with the plane.

once again Tom, do I just add the coefficent of sf on to the answer or do I have to say:
Not sure what you mean by this. You treat the frictional force exactly the same as always. It's magnitude is μ*N, where N is the magnitude of the normal force.

a(y)=0 mu(s)N = mu(s)*mgsin(theta) and that is an additional force since it is moving with the sphere. I understand that the friction is what is required for the sphere to not slip.
I'm not sure of what you mean by a(y), but here are some notes:

You solve it like any other inclined plane problem, but now you have rolling involved.

The static frictional force acts parallel to the incline, in the direction of motion. It's magnitude is μs times the normal force.

The normal force acts perpendcular to the incline, as usual.

The gravitational force acts straight down.

That's it--those are the only forces, so write down ΣF=ma, where a points down the incline (down, because the ball is decelerating.

Finally, you can sum the torques about the center of mass and set that equal to Iα.

That will give you the system of equations you need.
 
Last edited:
39
0
guess Im just a little rusty on incline plane problems. I dont have my book right here in front of me. But I am going to head out now. Any other tips that you can give me on how to solve all 3 of the questions?

if not, thanks for your help

M2k
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,475
20
Not just yet. Try to write out the following 3 equations from the free body diagram:

1. ΣFparallel=ma
2. ΣFperpendicular=0
3. Στ=Iα

and we'll go from there.
 

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