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Okay, let's assume that we have an incline which is at say, an angle $\theta$, not too big.

Now, we have a a bunch of different objects with all the same radius and the same mass. Let's say that the objects are a sphere, hoop, shell, and disk, so clearly the moment of inertias are different.

In addition, the incline is so rough i.e. $\mu$ is so big that all the objects will roll down the incline without slipping.

If the force acting on the objects if always $$\Sigma F_{net} = mg\sin\theta - mg\cos\theta\mu = ma$$ by Newton's Second Law, wouldn't the acceleration be the same in all cases?

I know there's something wrong with my reasoning, but I can't figure out what. Any help would be greatly appreciated.

EDIT

The reason my reasoning is wrong is because friction is not always $mg\cos\theta\mu$ but instead this expression is for the maximum possible value of friction, which is not what the friction in the force equation is supposed to represent. Thanks to everybody for helping me understand this :)

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  • $\begingroup$ By following your argument, if the friction is so big that the objects are not slipping, when you calculate $\vec F_{net}$ you will see that they should also not accelerate at all because the net force acting on them is $0$. $\endgroup$ – onurcanbektas Feb 3 '18 at 4:27
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You are considering the $F_{\rm friction}$ to have a constant value of $\mu mg\cos\theta$ which is wrong. The maximum value of friction is given by $\mu N=\mu mg\cos\theta$ ,but that does not mean that the value of $F_{\rm friction}$ is $\mu N=\mu mg\cos\theta$. It can be any value smaller than it. Since you are telling that $\mu$ is very big so we can safely assume that this max value is not reached and slipping doesn't occur.

the remaining calculations are this:- enter image description here

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  • $\begingroup$ Ohhhh I get it, so my equation for friction is wrong. I see. Totally get it :) Thanks. $\endgroup$ – NL628 Feb 4 '18 at 3:09
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The friction force can be less than what you wrote in your equation, because there is no slipping. Imagine that your equation is correct for some body and some value of the friction coefficient, then it will not be correct for the same body and a higher value of the friction coefficient.

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The problem you are having is that those so called "laws" are only for linear motions, and if you apply them to this particular case you will see that (as I have written as comment)

By following your argument, if the friction is so big that the objects are not slipping, when you calculate $F⃗_{net}$ you will see that they should also not accelerate at all because the net force acting on them is $0$.

and this does not result in any contradiction, it just tells us that the objects are not going to any linear motion.

Now, what rotational motion ?

If you google it as "Newton's law for rotational motion", you will see some "law" for rotational motion similar to "laws" for linear motion.

And if you apply them to this case, you will again see that the objects has to move by rotating.


I'm little confused since the title and the main body of your question does not match, so I'm going to answer to the title also.

Moment of inertia in rotational motion is equivalent to the mass in linear motion, i.e given a torque/force both resist to the change in angular/linear velocity (i.e angular/linear acceleration).

In particular, since all the object in your example have the same mass (and for a fixed origin) if we assume that the torque acting on the object are the same, then the bigger moment of inertia is, the less the angular acceleration of the object will be.

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