Rolling without slipping down an inclined plane

[Huzaifa]

Homework Statement

The acceleration and velocity of a body rolling down without slipping on a frictionless inclined plane

Relevant Equations

$$
a=\dfrac{mg\sin \theta }{m+\dfrac{I}{r^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}} \cdots(1)
$$

$$
v=\sqrt{\dfrac{2mgh}{m+\dfrac{I}{r^{2}}}}=\sqrt{\dfrac{2gh}{1+\dfrac{K^{2}}{r^{2}}}}=\sqrt{\dfrac{2gs\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}} \cdots (2)
$$

The acceleration and velocity of a body rolling down without slipping on a frictionless inclined plane are given by

$$
a=\dfrac{mg\sin \theta }{m+\dfrac{I}{r^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}} \cdots(1)
$$

$$
v=\sqrt{\dfrac{2mgh}{m+\dfrac{I}{r^{2}}}}=\sqrt{\dfrac{2gh}{1+\dfrac{K^{2}}{r^{2}}}}=\sqrt{\dfrac{2gs\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}} \cdots (2)
$$

Here, $K$ is the radius of gyration, $m$ is the mass of the body, $r$ is the radius of the body, $\theta$ is the inclination of the plane, $h$ is the height of the slope, and $s$ is the length of the slope.

I am not able to derive the acceleration $a$ and velocity $v$. Please help me.

I derived time $t$ by dividing (2) by (1), but I am not able to derive (1) and (2).

$$
t=\dfrac{\sqrt{\dfrac{2gs\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}}}{\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}}=\sqrt{\dfrac{2s\left( 1+\dfrac{K^{2}}{r^{2}}\right) }{g\sin \theta }}
$$

 

[Huzaifa]

Here is the FBD of the rolling body

 

[Huzaifa]

I have tried to derive acceleration and velocity but I am not able to. I took the resultant of linear/translational acceleration/velocity and angular/centripetal/rotational acceleration-and-velocity, and used them to work out how Newton's second law of motion works for this case. Please help me.

 

[berkeman]

Sorry, just to clarify -- how can you roll without slipping while accelerating down a frictionless inclined plane?

 

[Huzaifa]

Sorry, just to clarify -- how can you roll without slipping while accelerating down a frictionless inclined plane?

I am not sure if friction is present or not. It is not given if friction is present or not. The inclined plane may have friction. I also realized that rolling without slipping is not possible in a frictionless inclined plane. But I think the inclined plane is frictionless in this case.

 

[vela]

I am not sure if friction is present or not. It is not given if friction is present or not. The inclined plane may have friction. I also realized that rolling without slipping is not possible in a frictionless inclined plane. But I think the inclined plane is frictionless in this case.

You can't have it both ways. Either the inclined plane is frictionless, in which case the ball can't roll down without slipping, or the ball rolls without slipping, which means friction must be present. Which is it?

 

[haruspex]

I am not sure if friction is present or not.

Then you should realize it must be present in order for the ball to roll faster instead of slipping.
But I am also unclear on what you are trying to do.
If you are trying to obtain equations (1) and (2) from first principles, you need to start with force and torque equations, but in post #3 you make no mention of forces.
Please post the work you have done.

 

[Huzaifa]

You can't have it both ways. Either the inclined plane is frictionless, in which case the ball can't roll down without slipping, or the ball rolls without slipping, which means friction must be present. Which is it?

Friction is PRESENT. It is not frictionless inclined plane. I am sorry for the confusion. But the ball is rolling without slipping/skidding.

 

[Huzaifa]

I found the derivation of acceleration from this post from Physics.SE.
$$a=\frac{g\sin\theta}{1+\frac{k^2}{r^2}}$$
And according to this post, "the acceleration is independent of the friction for an object rolling down a plane".

 

[haruspex]

I found the derivation of acceleration from this post from Physics.SE.
$$a=\frac{g\sin\theta}{1+\frac{k^2}{r^2}}$$

That's an equation, not a derivation.
You still haven't explained what you are trying to do. Are you trying to derive equations (1) and (2) from Newton's laws? If so, please post at least one of your own equations in that attempt.

And according to this post, "the acceleration is independent of the friction for an object rolling down a plane".

If the coefficient of static friction is enough to ensure rolling (greater than the tan() of the slope) then it doesn't matter what its exact value is. I hope that is what was meant.
It is not true that it is independent of the frictional force.

 

[Huzaifa]

You still haven't explained what you are trying to do. Are you trying to derive equations (1) and (2) from Newton's laws?

Yes, I trying to derive equations (1) and (2) as mentioned above.

If so, please post at least one of your own equations in that attempt.

I have drawn the FBD for the rolling ball and I am trying to derive equations for the rolling ball from the Newton's laws like F=ma, f=μN, etc. The equations are:
$$a=\dfrac{mg\sin \theta }{m+\dfrac{I}{r^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}} \cdots(1)$$
$$v=\sqrt{\dfrac{2mgh}{m+\dfrac{I}{r^{2}}}}=\sqrt{\dfrac{2gh}{1+\dfrac{K^{2}}{r^{2}}}}=\sqrt{\dfrac{2gs\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}} \cdots (2)$$

please post at least one of your own equations in that attempt.

I have no equations of my own, I am not Newton. But I know F=ma, f=μN, etc.

 

[haruspex]

Yes, I trying to derive equations (1) and (2) as mentioned above.

I have drawn the FBD for the rolling ball and I am trying to derive equations for the rolling ball from the Newton's laws like F=ma, f=μN, etc. The equations are:
$$a=\dfrac{mg\sin \theta }{m+\dfrac{I}{r^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}} \cdots(1)$$
$$v=\sqrt{\dfrac{2mgh}{m+\dfrac{I}{r^{2}}}}=\sqrt{\dfrac{2gh}{1+\dfrac{K^{2}}{r^{2}}}}=\sqrt{\dfrac{2gs\sin \theta }{1+\dfrac{K^{2}}{r^{2}}}} \cdots (2)$$

I have no equations of my own, I am not Newton. But I know F=ma, f=μN, etc.

An attempt would consist of using with Newton's laws in the context of the set up in the FBD so as to get equations specifically relating unknowns in the FBD.
You have not done that anywhere.

The motion you are interested in is parallel to the plane, so the first step is to identify forces which have a component parallel to the plane. Second step is to write expressions for those components, and the third is relate them to the acceleration via $\Sigma F=ma$.

 

[robphy]

In general, rolling (related to $\omega$) is independent of sliding (related to $v_{cm}$).
But for "rolling without slipping", there is a relation (a constraint) between them.

 

[phinds]

Friction is PRESENT. It is not frictionless inclined plane. I am sorry for the confusion. But the ball is rolling without slipping/skidding.

Well, you think maybe you should change the subject line of the thread? There is no ambiguity about

Rolling without slipping down a frictionless inclined plane​

 

[Huzaifa]

An attempt would consist of using with Newton's laws in the context of the set up in the FBD so as to get equations specifically relating unknowns in the FBD.

The motion you are interested in is parallel to the plane, so the first step is to identify forces which have a component parallel to the plane. Second step is to write expressions for those components, and the third is relate them to the acceleration via ΣF=ma.

$$\begin{aligned}mg\sin \theta -F=ma\\ \Rightarrow mg\sin \theta -\mu R=ma\\ \Rightarrow mg\sin \theta -\mu mg\cos \theta =ma\\ \Rightarrow mg\left( \sin \theta -\mu \cos \theta \right) =ma\\ \Rightarrow a=g\left( \sin \theta -\mu \cos \theta \right) \end{aligned}$$

I think I am missing torque in my equation.

 

[haruspex]

View attachment 290739
$$\begin{aligned}mg\sin \theta -F=ma\\ \Rightarrow mg\sin \theta -\mu R=ma\\ \Rightarrow mg\sin \theta -\mu mg\cos \theta =ma\\ \Rightarrow mg\left( \sin \theta -\mu \cos \theta \right) =ma\\ \Rightarrow a=g\left( \sin \theta -\mu \cos \theta \right) \end{aligned}$$

I think I am missing torque in my equation.

Yes, you are missing torque, perhaps because you have drawn F acting at the centre of the disc. Where does it act?
But there is also an error in the equations. You have assumed $F=\mu_sR$. In what circumstance would that be true? Does that apply here?

 

[Huzaifa]

 

[Huzaifa]

How did we get these three equation in the video?
$$\begin{aligned}mg\sin \theta -f=ma\cdots \left( 1\right) \\ f\cdot R=I\alpha \cdots \left( 2\right) \\ a=R\alpha \cdots \left( 3\right)\end{aligned}$$
I understand the first equation, but I am not able to understand the later two. What exactly is R here in equation 2&3?
$$\begin{aligned}f\cdot R=I\alpha \cdots \left( 2\right) \\ a=R\alpha \cdots \left( 3\right)\end{aligned}$$
Please help me with this. I have been struggling for the last two months.

 

[haruspex]

How did we get these three equation in the video?
$$\begin{aligned}mg\sin \theta -f=ma\cdots \left( 1\right) \\ f\cdot R=I\alpha \cdots \left( 2\right) \\ a=R\alpha \cdots \left( 3\right)\end{aligned}$$
I understand the first equation, but I am not able to understand the later two. What exactly is R here in equation 2&3?
$$\begin{aligned}f\cdot R=I\alpha \cdots \left( 2\right) \\ a=R\alpha \cdots \left( 3\right)\end{aligned}$$
Please help me with this. I have been struggling for the last two months.

R is the radius (your r). Since f acts at the point where the wheel touches the slope, and acts directly up the slope, its perpendicular distance from the centre of the wheel is R. Hence its torque about the centre of the wheel is fR. Thus $fR=I\alpha$, where I is the moment of inertia of the wheel about its centre and $\alpha$ is its angular acceleration.
If a rolling wheel radius r turns through an angle $\theta$ then it moves distance $s=r\theta$ along the surface. Differentiating, we find, successively, $v=\dot s=r\dot\theta=r\omega$ and $a=\dot v=r\dot \omega=r\alpha$.