Rotational Motion/Projectile motion of a ball on a circular ramp.

1. Jun 23, 2011

andorrak

1. The problem statement, all variables and given/known data

A small sphere of radius r0 = 1.4 cm rolls without slipping on the track shown in the figure whose radius is R0 = 40.0 cm. The sphere starts rolling at a height R0 above the bottom of the track. Assume that it leaves the track after passing through an angle of 135° as shown.

Picture: http://imgur.com/7YLlJ

It asks for speed at the end of the ramp. and what is d.

2. Relevant equations

I know all the basic equations. I am just confused how to get the height at the end of the ramp. ie. heres my equation.

E0=mgh or m(9.8)(.4)

E1=.5mv2+.5Iw2 +mgh or

I replace "I" with the moment of inertia for a perfect sphere which is 2/5mr2

.5mv2+.5(2/5mr2)v2+m(9.8)(h)

E1 is the energy at the end of the ramp. But i cannot get h which is that height at the end of the ramp.

I figure the angle of that circle must be 45 degrees. So that arc length is .31m. Since the Radius is 40cm. Arc Length = angle times radius.

But i cannot get the height. If i get the height i can solve for the speed, then the get d.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 23, 2011

ehild

Try to draw the unknown height into the picture before you say 'I can not get h'. See attachment.

ehild

Attached Files:

• ramp.JPG
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3. Jun 23, 2011

andorrak

how can you say that at the point of the circle will make a full radius above that point?

Yea you absolutely cant. the triangle would have two lengths of radius 40 cm repeated on the hypotenuse and the sides.

4. Jun 23, 2011

ehild

I certainly have not said that.

ehild

5. Jun 23, 2011

andorrak

then i have no idea what is going on geometrically

6. Jun 23, 2011

ehild

Examine the triangle. What is the length of the hypotenuse?
What is its angle?

ehild

7. Jun 23, 2011

andorrak

haha i feel like an idiot now. math does that to you.

but how do i get d. i think my kinematics get me the distance from the point it shot off the ramp. i cant make a triangle underneath to get it.

8. Jun 23, 2011

andorrak

But my answers dont seem to be right. here was my math.

I had h being .1171 m.

so the velocity turned out to be 1.76 m/s which is wrong.

I redid the math as I was looking at my work and i dont know what is wrong just wanted to check that with you.

I get:

9.8*.4=.5v2+(1.4^2)(.5)(.4)v2+9.8*.1171

2.77242=.892v2

v=1.76

9. Jun 24, 2011

ehild

You count the translational KE twice.
Edit: and included Ro into the rotational energy, why?

ehild

Last edited: Jun 24, 2011
10. Jun 24, 2011

flyingpig

11. Jun 24, 2011

ehild

The centripetal force does not "act" it is just the resultant of other forces. You might have thought of the normal force. But it is normal to the trajectory, so its work is zero.

ehild

12. Jun 24, 2011

flyingpig

Is it possible to solve using torque and other newton's laws?

Feel abd for hijacking (I won't show the steps), but I got roughly 1.98m/s

13. Jun 24, 2011

ehild

You can solve it by other methods than conservation of energy, but it seems the simplest. Your result is about correct.

ehild

14. Jun 24, 2011

andorrak

You count the translational KE twice.
Edit: and included Ro into the rotational energy, why?

ehild

I thought I am supposed to count both the kinetic energy of the ball and the rotational energy, which I thought accounts for the energy total. and what is Ro?

15. Jun 24, 2011

ehild

Yes, but what is the rotational energy? Ro is the radius of the track in the figure, 0.4 m.

ehild

16. Jun 24, 2011

andorrak

I thought the rotational energy is the one that involves the sphere. The 1.4cm

17. Jun 24, 2011

ehild

So 1.4 was cm and the radius of the sphere! Now I understand where is it from, and what is 0.4. But: The rotational energy is not 0.5 I v^2, but 0.5 I ω^2 (ω is the angular velocity).

ehild

18. Jun 24, 2011

andorrak

I use 0.4 m in the beginning of the equation to represent the PE of the initial state before it begins to roll. I know that the rolling energy is .5Iw^2. But i replace I with 2/5r^2 which is the moment of inertia for a sphere. so the .4 here might be the confusion u are talking about.

hence I get for the rolling energy of the sphere:

.5*.4*(1.4/100)^2 *v^2

I also realized ive been using 1.4 METERS as the radius. Derp. but i still do not get the answer. With the 1.4 cm use I get the velocity as 2.35 m/s. Which i know is wrong. Any suggestions Ehild?

19. Jun 24, 2011

ehild

You can not use v instead of w. v is the velocity of the CM, w is the angular speed of rotation. How are they related in case of pure rolling?
Check the dimension of your formula. Disregarding mass, it has the unit of m^4/s^2 instead of m^2/s^2.

ehild

20. Jun 24, 2011

andorrak

Of course i know that dude.

but remember Omega = V/R? so the rotational energy is .5IW^2. I was simply canceling out R ahead of time.

I= .4R^2 and we have (V/R)^2. the R's cancel out So i just get energy.