Energy, Force, and Inertia Problem. Ball down ramp to loop to loop

[rschaefer2]

Homework Statement

A solid sphere of mass m and radius r rolls without slipping along the track shown below. It starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, much larger than r. (Consider up and to the right to be the positive directions for y and x respectively)What are the force components on the sphere at the point P if h = 3R? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

Homework Equations

I_sphere=(2/5)mr²
K=.5mv²
K=.5Iω²
PE=mgh
a_c=v²/R

The Attempt at a Solution

F_x
First, solving for v²
mg(3R)=.5mv²+(.5)(2/5)(mr²)(v²/r²)
mg2R=.7mv²+mgR
v²=20gR/7

Centripetal acceleration is the acceleration in the x direction
a=v²/R=20gR/7R
F_x=ma=-20mg/7 <=== Correct!

F_y
Force in y direction is only the force from gravity
F=ma
F_y=-mg <=== Wrong!

please help! What am I missing?

 

[TSny]

Hello, and welcome!

Note that you are measuring heights from the "floor". Does point P have a height?

EDIT:

mg(3R)=.5mv²+(.5)(2/5)(mr²)(v²/r²)
mg2R=.7mv²
v²=20gR/7

OK, I saw that you left out the potential energy at P in the first equation, but it looks like you took care of it in the second equation.

 

[TSny]

I don't see anything wrong with your answer for F_y. Wait! Is the force of gravity the only force with a y component acting on the ball at P?

 

[rschaefer2]

Well, i guess there is friction to? But there wasn't any coefficients given for kinetic friction

 

[TSny]

Right. Is the friction kinetic or static?

 

[rschaefer2]

Actually, static correct? Because the object is rolling.

 

[TSny]

That's right- rolling without slipping. So, even if they did give you a coefficient of static friction it wouldn't be of much help. The static coefficient only let's you find the maximum force of static friction, and there is no reason why the static friction would need to be at its maximum value at P.

So, you'll need to find the friction force some other way. Will the ball have any angular acceleration at P?

 

[rschaefer2]

The ball itself? Well angular acceleration will be the derivative of ω with respect to s? EDIT: s as in seconds, from g

 

[TSny]

That's a correct statement if s denotes time. [Edit: Ah, I see that you are using s for time!] But it's not of much help.

Do you think the ball will be speeding up or slowing down at P?

 

[rschaefer2]

Definitely slowing down, because v will be decreasing

 

[TSny]

Will ω also be decreasing?

 

[rschaefer2]

Yes, because v=ωr

 

[TSny]

Good. So, looks like you have an unknown y-component of acceleration, an unknown angular acceleration, and an unknown friction force. You'll need three independent equations that relate these unknowns. How many equations can you think of?

 

[rschaefer2]

1) \sumτ=I(alpha)
2)v=ωr
3)f_s=μF_n
4) alpha=a_T/r
5)a_T=(-mg-f_s)m
F_n= -20mg/7
I=2/5mv²/r²

I can't seem to relate it using these, I'm probably missing one of the equations.

 

[TSny]

Oooo. Shotgun approach. Need to narrow the field. Eq. #1 looks like a keeper. #4 is also good except the subscipt T is suspicious. What does a_T stand for?

Finally, you'll need another equation that relates at least two of the three unknowns: a_y, \alpha, and the friction force. (Hint: We don't want to upset Newton - which is easy to do.)
[EDIT: your #5 equation looks like it might be the third equation. Not sure the signs are correct though and there appears to be another mistake in it also.]

 

[rschaefer2]

a_t=a_y

Okay, thanks! I'll try to work them out.

 

[TSny]

Ok, good. (I was going to have to quit for now anyway - it's late and I need my beauty rest.)

 

[rschaefer2]

My attempt:

Equation 1 expands to:
mg+f_s=(2/5)mr²(a_y/r)

plugging the right side of this equation into equation 3:

a_y=(2/5)mra_y/m

and this is where i get stuck, because the acceleration terms cancel out.

 

[TSny]

Equation 1 expands to:
mg+f_s=(2/5)mr²(a_y/r)

What you have written on left side doesn't represent the type of quantity you wrote earlier on the left side of equation 1. (The dimensions of what you have on the left don't match the dimensions you have on the right.)

Also, it might help to think about the direction of the friction force. If the angular speed of the ball is slowing down at P, what must be the direction of the net torque on the ball (clockwise or counterclockwise)? What direction does the friction force need to act in order to produce that direction of torque about the center of the ball?

 

[rschaefer2]

EDIT: From previous, I've changed torque to only static friction in the positive direction, as mg acts through the axis. Also, the friction force needs to counteract the clockwise movement, therefore needing to be positive (counterclockwise).

So I'm pretty sure I've worked this out correctly. I just want to double check, as this is my last attempt for points. Thanks!

Equations:
1) \sumτ=Iα
f_sr=Iα
α=f_sr/I

2)α=a_y/r
f_sr/I=a_y/r
a_y=f_sr²/I

3)ma_y=(-mg+f_s)
mf_sr²/((2/5)mr²)=-mg+f_s
5/2f_s-f_s=-mg
3/2f_s=-mg
f_s=-2/3mg

Plug back into 1)
α=(-2/3)mgr/I
α=(-10g/6r)

Plug into 2)
(-10g/6r)=a_yr
a_y=(-10g/6)

F_y=ma_y
F_y=(-10mg/6)

 

[TSny]

1) \sumτ=Iα
f_sr=Iα

2)α=a_y/r

3)ma_y=(-mg+f_s)

Your equation (1) indicates that you are choosing the positive direction for rotation to be the direction of the torque produced by f_s. Is that clockwise or counterclockwise?

Your equation (3) indicates that you are choosing positive y in the upward direction.

Now, if a_y is positive, is the angular acceleration α positive or negative according to your conventions adopted in (1)?
Equation (2) implies that α is positive when a_y is positive. Is that correct?

Dang signs

 

[rschaefer2]

Rolling up the ramp, the ball is rotating clockwise. Since ω is decreasing, the torque acting on the sphere must be in the opposite direction, counter clockwise.

So the positive direction of rotation is counterclockwise

If ay is positive, then the angular speed should also increase, therefore it should rotate clockwise quicker. So equation two should be a_y=-αr?

 

[TSny]

Rolling up the ramp, the ball is rotating clockwise. Since ω is decreasing, the torque acting on the sphere must be in the opposite direction, counter clockwise.

So the positive direction of rotation is counterclockwise

If ay is positive, then the angular speed should also increase, therefore it should rotate clockwise quicker. So equation two should be a_y=-αr?

Yes, I believe so. Note that you got a negative value of f_s which would indicate that the friction force is downward. But you know that it is upward. So, see how it works out with a_y=-αr

 

[rschaefer2]

Awesome, thank you! my new answer is (-10/14)mg, with the correct signs. Any confirmation?

 

[TSny]

That's what I got, too. You might want to reduce the fraction. Good!