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Old Mar 9th 2017, 02:38 PM   #1
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Question Is this correct?

My formula for a ball rolling down a ramp:

(Weight)sin(angle) = a
(Weight)cos(angle) = b
(Friction) times b = c
a - c = d
d divided by mass = e
e + (5 x 9.8 x sin(angle) ) / 2 = acceleration

Is this correct?

Last edited by Zman; Mar 9th 2017 at 02:47 PM. Reason: Typo
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Old Mar 9th 2017, 04:41 PM   #2
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Originally Posted by Zman View Post
My formula for a ball rolling down a ramp:

(Weight)sin(angle) = a
(Weight)cos(angle) = b
(Friction) times b = c
a - c = d
d divided by mass = e
Correct so far - 'e' is the object's acceleration down the ramp. I would point out however that these formulas are for an object that is sliding down a ramp, not a ball rolling.

Originally Posted by Zman View Post
e + (5 x 9.8 x sin(angle) ) / 2 = acceleration
Is this correct?
What are you attempting to do here? If you draw a free body diagram of the forces acting on the sliding object you have accounted for everything acting along the direction of the incline: force of gravity (your 'a') and friction ('c'). It seems you are trying to account for some third force here - what would that be?
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Old Mar 9th 2017, 07:24 PM   #3
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e + (5 x 9.8 x sin(angle) ) / 2 = acceleration

The line above is to take accout of the rolling. Another way that I've heard for rolling is the line minus "e +" and get ride of the sliding formula. Is that one correct or do I stick with my original? Or is there a third formula all together?
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Old Mar 10th 2017, 02:23 AM   #4
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So you are introducing a term:

r=(5 x 9.8 x sin(angle) ) / 2

and then:

(acceleration) =e + r

I don't see how a rolling term comes into the acceleration.
It might come into the friction term, but I can't see how it can fit where you have it.
As I see it you should stop at:

(acceleration) = e
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Old Mar 10th 2017, 05:29 AM   #5
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E is the acceleration for the ball sliding. "r" as you put it is supposed to take account the rolling. The rolling has to be part because some energy is lost to making the ball spin. Again another way I've heard for rolling and not sliding is just take "r" without "e" and use it as acceleration.

Last edited by Zman; Mar 10th 2017 at 05:37 AM.
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Old Mar 12th 2017, 10:59 AM   #6
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I have been thinking about this, and while I suppose there could be a term due to the rotational inertia of the ball, it would be very small.
Certainly a lot less than your friction term, indeed rolling is used to reduce friction, this would be pointless if it introduced a significant deceleration term.

I think that for all practical problems any rotational term would be negligible,
indeed for most purposes the frictional term can be ignored.

Lastly, even if you were to include a term due to the rotation, I cannot see how the equation you have given could be anywhere near correct.
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Old Mar 12th 2017, 07:42 PM   #7
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What would be correct?
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Old Mar 12th 2017, 08:23 PM   #8
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I think I get what you're trying to say, but I don't believe it's quite right. First, your 'r' term comes from an equation for torque = moment of inertia times angular acceleration:

$\displaystyle T = I \alpha $

For a solid sphere rolling on an incline it may seem that $\displaystyle T = mg \sin (\theta) R, \ \ I = (2/5)mR^2$ and [math] \alpha = \frac a R[/math, which yields:

$\displaystyle a = \frac 5 2 g \sin (\theta) $

which is what you added to your "e" sliding acceleration term. The problem with this is that it ignores the fact that the ball must accelerate in translation as well as rotation. You can't simply add this to the acceleration you calculated for a sliding object - in effect this is saying that a rolling object can accelerate faster than a sliding object, even if there is no friction, which is clearly not right. Note also that if sin(theta) = 1 this would say that the ball accelerates at a rate faster than g, which again is not right.

A better way is to apply energy methods:

$\displaystyle \Delta PE = - \Delta KE $

For the rolling ball on an incline:

$\displaystyle mgd \sin(\theta) = \frac 1 2 mv^2 + \frac 1 2 I \omega ^2 $

where 'd' = displacement along the incline. If you work it though you should find that:

$\displaystyle v = \sqrt {\frac {10} 7 gd \sin \theta} $

Note that the friction force does not contribute to the analysis, as long as the ball is rolling and not sliding. Also note that this assume that $\displaystyle v_0 = 0$. You can then determine distance as a function of time:

$\displaystyle v = \frac {dx}{dt} = \sqrt { \frac {10} 7 g \sin \theta x}$

(Note that here I have switched from 'd' to 'x' for the distance the ball has rolled down the ramp.)

$\displaystyle \int \left( \sqrt {\frac {10} 7 g \sin \theta}\ \right) dt= \int x ^{-1/2} dx $

$\displaystyle \left( \sqrt {\frac {10} 7 g \sin \theta}\ \right) t = 2 \sqrt x $

$\displaystyle x = \frac 5 {14} g \sin \theta t^2 $

One other point: all this assumes the ball is rolling down the incline. But if the friction force is small enough there isn't enough torque provided by friction to get the ball rolling and it will slide rather than roll. So we need to check whether that is the case. In order to roll, the torque from friction must be at least equal to the torque required to get the ball spinning at the required rate:

$\displaystyle T = \mu mg \cos \theta \ge I \alpha$

We can find alpha from the second derivative of the equations we derived above: $\displaystyle a = \frac {d^2x}{dt^2} = \frac 5 7 g \sin \theta$, and $\displaystyle \alpha = \frac a {r}$. Plug this in, along with the known value for I:


$\displaystyle \mu mg \cos \theta r \ge \frac 2 5 mr^2 \frac 5 7 \frac {g \sin \theta} r $

which leads to:

$\displaystyle \mu \ge \frac 2 7 \tan \theta$

If mu is less than this value, the ball will slide while slowly beginning to rotate, which complicates the analysis. Post back if you want to go deeper into what happens under this condition.

Hope this helps.

Last edited by ChipB; Mar 13th 2017 at 02:17 PM.
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