Calculating Gravity from Inclined plane

[Fat_Squirrel]

Homework Statement

http://i.imgur.com/Osn6pKO.png

2m long Inclined plane. Height at top of the ramp is 6.1cm
We are given distances of 0.5, 1, 1.5 2.0m and their times (close to 2.5s, 3.5s ,4.25, 5s).

Need to calculate g.

Homework Equations

d=1/2at^2
g=a/(sinθ/(1+I/mr^2)) (where sinθ=0.061/2)

The Attempt at a Solution

So I plotted all the data in excel. Graphed time squared against distance and excel tells me the gradient of that slope is 0.0823, so a is twice that, 0.1646.

If I use g=a/sin(theta), 0.1646/(0.061/2) = 5.39 .. pretty far off 9.8.

So I assume I have to use the additional info they give me (the diameter of the ball and the V shaped ramp) and add in the effect of the ball rolling. But I'm not sure what formua to use. I thought I/mr^2 for a sphere is 2/5, but then my answer is even further from what I know to be g.

Suggestions?

 

[Fat_Squirrel]

Actually my answer gets closer, but I'm looking at g=7.58 ?? Seems a fair way off still to me. What am I doing wrong?

 

[haruspex]

It is not the same as a ball rolling down a flat plank. Please post the algebraic details of your calculation (don't plug in numbers).

 

[Fat_Squirrel]

It is not the same as a ball rolling down a flat plank. Please post the algebraic details of your calculation (don't plug in numbers).

I was going with a=5/7*g*sin θ, so g = 7a / 5*sinθ

How do I account for it not being equivalent to a flat plane ?

 

[Fat_Squirrel]

It is not the same as a ball rolling down a flat plank. Please post the algebraic details of your calculation (don't plug in numbers).

Or are you asking how I got the answer for a ? If so, I took average of the times, squared these times and plotted it against distance.
Excel then gave me line of best fit with gradient = 0.0823

As d=1/2at^2, we need to double the gradient to get a, so 0.1646.

 

[haruspex]

I was going with a=5/7*g*sin θ, so g = 7a / 5*sinθ

How do I account for it not being equivalent to a flat plane ?

One way of arriving at the 5/7, for a flat plank, is to consider the rolling ball as rotating about its point of contact with the plank. Applying the parallel axis theorem, $mr^2+\frac 25 mr^2=\frac 75 mr^2$.
Where is the instantaneous axis of rotation in the present case? What does the parallel axis theorem give for that?

 

[Fat_Squirrel]

Where is the instantaneous axis of rotation in the present case?

Honestly, no idea. Purely guessing I'd say its the midpoint on the chord between the two points of contact which I calculated to be very nearly 10.41mm from the centre of the ball.

Even if that's correct, I have even less of an idea what I would do with this figure.

 

[Fat_Squirrel]

Honestly, no idea. Purely guessing I'd say its the midpoint on the chord between the two points of contact which I calculated to be very nearly 10.41mm from the centre of the ball.

Even if that's correct, I have even less of an idea what I would do with this figure.

1 + 10.4/15 = 1.69333 if I multiply that by my original a/sinθ , I get 9.14. My closest answer yet. But based on nothing but fact I know somewhat what I need to multiple this answer by to get close to g.

Tried watching some youtube videos, sadly they break into Indian at a moment's notice :(

 

[dean barry]

starting from scratch:
find the effective rolling radius of the ball in the groove.
(the distance between the ball centre and the centre of the line between the contact points)
Imagine the ball moving at a constant (arbitrary) 10m/s in the groove, and
calculate the linear KE and the rotating KE at this speed.
Then:
Calculate the effective mass (em) of the ball from:
em = ( 1 + ( rotating KE / linear KE ) * actual mass of the ball
The acceleration of the ball (a) should be :
a = f / em
(you know a from your data)
(f is the driving force due to gravity = m * g * sine ( incline angle)
jumble to isolate g

 

[Fat_Squirrel]

starting from scratch:

Calculate the effective mass (em) of the ball from:
em = ( 1 + ( rotating KE / linear KE ) * actual mass of the ball

We aren't given the mass of the ball anywhere sadly. I must have read the question a half dozen times to make sure I didn't miss it.

 

[dean barry]

i set this up on excel with a fixed rolling radius and variable masses (solid sphere assumed in all cases) and the acceleration varies with the mass
i can't see how to solve it.

 

[haruspex]

starting from scratch:
find the effective rolling radius of the ball in the groove.
(the distance between the ball centre and the centre of the line between the contact points)
Imagine the ball moving at a constant (arbitrary) 10m/s in the groove, and
calculate the linear KE and the rotating KE at this speed.
Then:
Calculate the effective mass (em) of the ball from:
em = ( 1 + ( rotating KE / linear KE ) * actual mass of the ball
The acceleration of the ball (a) should be :
a = f / em
(you know a from your data)
(f is the driving force due to gravity = m * g * sine ( incline angle)
jumble to isolate g

If you get your parentheses right you will discover mass cancels out.

 

[haruspex]

Purely guessing I'd say its the midpoint on the chord between the two points of contact which I calculated to be very nearly 10.41mm from the centre of the ball.

Yes, the axis is the chord joining the points of contact.
But this doesn't just change the effective inertia. It also changes the relationship between the linear velocity and the angular velocity.
Suppose the chord is distance x from the ball's centre (please avoid plugging in numbers for now). What is the relationship between v and $\omega$? What is the moment about the axis? What does that give you for the acceleration?

 

[Fat_Squirrel]

Yes, the axis is the chord joining the points of contact.
But this doesn't just change the effective inertia. It also changes the relationship between the linear velocity and the angular velocity.
Suppose the chord is distance x from the ball's centre (please avoid plugging in numbers for now). What is the relationship between v and $\omega$? What is the moment about the axis? What does that give you for the acceleration?

Avoid plugging numbers in? I don't even know what formula I'm supposed to be using. I have no idea what you're talking about. Sorry, but I give up.
I'm going to need something beyond these cryptic clues. This is my first month of physics and this is all well above my head.

 

[haruspex]

Avoid plugging numbers in? I don't even know what formula I'm supposed to be using. I have no idea what you're talking about. Sorry, but I give up.
I'm going to need something beyond these cryptic clues. This is my first month of physics and this is all well above my head.

As moment of inertia problems go, this is somewhat advanced.
Draw yourself a side view. Say the ball has radius R, and the centre of the ball is distance A from the chord joining the points of contact with the ramp. If at some instant the ball is rotating at rate $\omega$, what will be its linear speed down the ramp? What will its total KE be (linear plus rotational)?

 

[Fat_Squirrel]

As moment of inertia problems go, this is somewhat advanced.
Draw yourself a side view. Say the ball has radius R, and the centre of the ball is distance A from the chord joining the points of contact with the ramp. If at some instant the ball is rotating at rate $\omega$, what will be its linear speed down the ramp? What will its total KE be (linear plus rotational)?

Appreciate the assistance. But I've already spent WAY too long on this problem. It's only worth a single bonus mark. It's not worth my mental heath to continue it further. I'll just wait for the answer and live without that mark.

Thanks for the help though.

 

[Fat_Squirrel]

Yes, the axis is the chord joining the points of contact.
But this doesn't just change the effective inertia. It also changes the relationship between the linear velocity and the angular velocity.
Suppose the chord is distance x from the ball's centre (please avoid plugging in numbers for now). What is the relationship between v and $\omega$? What is the moment about the axis? What does that give you for the acceleration?

v= rω ??
Moment about the axis? No idea. I_com + Mr^2 ? Honestly, what am I missing?

Can you maybe explain this like I'm 5? I can't find anything discussing these ideas that don't include mass, and I can't see how to remove the mass from the equation, and I can't see how we calculate the inertia at a point off the centre without using mass I don't have.

How do we use the parrallel axis theorum withhout mass? i don't see how we can remove it from what Dean Barry suggested. And everything that derives these equations always uses mass. Textbook, youtube vids, physics forums aren't helping.

Can you please give me a big nudge in the right direction?

 

[haruspex]

v= rω ??

No. Did you draw yourself a diagram as I suggested? You should have a circle radius r representing the ball, and a sloped line that cuts through it representing the line of contact with the ramp. At its closest point, the ramp is distance x from the centre of the ball. It might help to draw another circle concentric with the first, radius x.
As the ball rolls down the ramp, if it turns an angle theta, how far has it moved down the ramp?

Moment about the axis? No idea. I_com + Mr^2 ?

Right general formula. In that formula, r represents the distance from the mass centre to the instantaneous centre of rotation. In the present case, the instantaneous centre of rotation is the chord, so its distance x from the mass centre.
In terms of m, r and x, what is I_com?

I can't find anything discussing these ideas that don't include mass

Just put the mass as an unknown m. You'll see what happens later.

 

[Fat_Squirrel]

In terms of m, r and x, what is I_com?

I_com=2/5mr^2 ?
So I = 2/5mr^2 + mx^2 ?

 

[haruspex]

I_com=2/5mr^2 ?
So I = 2/5mr^2 + mx^2 ?

Good. Now what about the relationship between v, r, x and omega?

 

[Fat_Squirrel]

Good. Now what about the relationship between v, r, x and omega?

v= (r-x)ω ?

 

[haruspex]

v= (r-x)ω ?

No, but maybe you're defining x differently from how I defined it. I put x as the distance from the ball's centre to the plane containing the lines of contact with the ramp.

 

[Fat_Squirrel]

ω

No, but maybe you're defining x differently from how I defined it. I put x as the distance from the ball's centre to the plane containing the lines of contact with the ramp.

So just v=xω then

 

[haruspex]

ω

So just v=xω then

Yes.

 

[Fat_Squirrel]

Yes.

What are we working towards here, total Kinetic energy ? Sorry, it's just not clicking.

What's our end goal here? If I know what we're working towards that might help me put the pieces into place.
We have a value for x, are we trying to work out ω, or looking for something to substite in place of v? Sorry, I've looked at so many formulas its making less and less sense.

Was the other person here posting about effective mass on point, or was that a red herring?

 

[haruspex]

What are we working towards here, total Kinetic energy ? Sorry, it's just not clicking.

What's our end goal here? If I know what we're working towards that might help me put the pieces into place.
We have a value for x, are we trying to work out ω, or looking for something to substite in place of v? Sorry, I've looked at so many formulas its making less and less sense.

Was the other person here posting about effective mass on point, or was that a red herring?

We are now in a position to work out what fraction of the KE goes into linear motion and what into rotational. That will give us the ralationship between distance moved down the ramp and linear velocity.

 

[Fat_Squirrel]

We are now in a position to work out what fraction of the KE goes into linear motion and what into rotational. That will give us the ralationship between distance moved down the ramp and linear velocity.

K_linear = 1/2mv^2
K_rotational = 2/5mr^2 + mx^2 ?

Total KE = 1/2mv^2 + 2/5mr^2 + mx^2 ??

So rotional/linear = (1/2 v^2) / (2/5r^2 + x^2) ??

?

 

[haruspex]

K_rotational = 2/5mr^2 + mx^2 ?

That's the moment of inertia about the instantaneous axis. What do you have to do with that to get the KE?

 

[Fat_Squirrel]

That's the moment of inertia about the instantaneous axis. What do you have to do with that to get the KE?

KE=1/2Iω^2

KE=1/2I(v^2/x^2) ??

 

[Fat_Squirrel]

KE=1/2Iω^2

KE=1/2*I*(v^2/x^2) ??

Which means our ratio should be 1 / (2/5 r^2/x^2 + 1) ??

 

[haruspex]

Which means our ratio should be 1 / (2/5 r^2/x^2 + 1) ??

Almost.
I'm afraid I may have confused you a little.
When dealing with bodies that are both rotating and moving linearly, there are two main ways of thinking of it:
- as a linear movement of the mass centre, plus a rotation about the mass centre, or
- as a rotation about the instantaneous centre of rotation.
With the first approach, the total KE is $\frac 12 m(v^2+\frac 25 r^2\omega^2) = \frac m2 (x^2+\frac 25 r^2)\omega^2$, exactly the result the second approach gives.
Hence the fraction of the total KE that is linear is $\frac{x^2}{x^2+\frac 25 r^2}$.
Can you see how to get from there to an expression for v as a function of the height loss?

 

[Fat_Squirrel]

1/[x^2 / (x^2 + 2/5 r^2)] * a * sinθ

so 1/[10.41^2 / (10.41^2 + 2/5 * 15^2) * 0.1646 * (0.061/2)] = 9.866

Answer looks good,how'd I go with this one?

 

[haruspex]

1/[x^2 / (x^2 + 2/5 r^2)] * a * sinθ

so 1/[10.41^2 / (10.41^2 + 2/5 * 15^2) * 0.1646 * (0.061/2)] = 9.866

Answer looks good,how'd I go with this one?

Yes, I believe that works.

 

[Fat_Squirrel]

Yes, I believe that works.

Once again, thanks for your help and patience. Will let you know if we got the answer my teacher expected.

 

[Fat_Squirrel]

Yes, I believe that works.

here's the solution they provided - https://youtu.be/IED3ujslfPY. I got a slightly different answer because I calculated acceleration differently.
I used values associated with t^2 and ran least squares on them to get my value for the slope, he used √d values - i thought they'd give same answer, but slightly different it seems.

Anyway if I use his value for acceleration I get the exact same answer he does. Thankfully my answer was within the margin of error. Thanks again.