# A ball (sphere) rotating along a moving incline (1 Viewer)

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#### titansarus

1. The problem statement, all variables and given/known data
We have a ball of mass $m$and radius $r$. it is placed on an incline (We don't know the angle of the incline, nor we do whether the angle is constant along the incline - maybe it is a curved incline) and then released. The COM of ball is $h$ meters above the incline at the initial time ($r$ is comparable to $h$ and is not negligible). Incline has mass $M$ which is also comparable to $m$ and is not very large, so incline will also move. We also have $\mu_s$ and $\mu_k$, static and kinetic friction coefficient. when the ball reaches the horizontal part of the incline, we want to find $v$ and $u$, velocity of ball and incline, relative to ground. ($I$ moment of inertia of Ball is known. $I= 2/3 mr^2$) (As you can see, the incline is somehow curved and the angle of slope is not constant)

3. The attempt at a solution

I can solve these type of question for a completely not moving incline with angle $\theta$. I write $\tau = I \alpha$ and $F = m a$ and $a = \alpha r$ and by knowing that at the first $\tau = r ~mg~cos\theta ~~\mu_k$, everything is solved easily. And also we can find a point where $v = \omega r$.

But In this question, we don't know anything about the incline except that it has a Mass and moves. And I don't know how to approach this problem.

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#### Chestermiller

Mentor
What is the exact statement of the problem?

#### titansarus

What is the exact statement of the problem?
The original question is not in English.
The first part is only some given numbers (eg. $\mu_s = 0.7$) which is actually not important in general solution and I didn't write them because I don't want numerical solution. After these numbers, It is said that we release a ball from height h from horizontal part of the incline. Assume that frictions coefficient are so large that the ball always roll. And the incline don't have friction with the ground. Calculate the $v$ and $u$ velocity of ball and incline, relative to ground at the time the ball reach the horizontal part of the incline. Just one thing I forgot to mention and that is this sentence: "BE CAUTIOUS about writing the condition of complete rolling (rolling without slipping). The relative velocity of two objects must be zero."

#### Chestermiller

Mentor
The original question is not in English.
The first part is only some given numbers (eg. $\mu_s = 0.7$) which is actually not important in general solution and I didn't write them because I don't want numerical solution. After these numbers, It is said that we release a ball from height h from horizontal part of the incline. Assume that frictions coefficient are so large that the ball always roll. And the incline don't have friction with the ground. Calculate the $v$ and $u$ velocity of ball and incline, relative to ground at the time the ball reach the horizontal part of the incline. Just one thing I forgot to mention and that is this sentence: "BE CAUTIOUS about writing the condition of complete rolling (rolling without slipping). The relative velocity of two objects must be zero."
Thank you very much. This fills in the important information that was missing from the original post (the part about no friction with the ground, and the part about no slippage at the contact between the ball and the slide).

So what are some of your thoughts on how to proceed?

#### titansarus

Thank you very much. This fills in the important information that was missing from the original post (the part about no friction with the ground, and the part about no slippage at the contact between the ball and the slide).

So what are some of your thoughts on how to proceed?
At first, I wanted to solve it like the usual inclines. But at this one, it seems we don't have a fixed angle and I can't write something like $cos \theta$. Maybe the conservation of energy could help here but I don't know how to write it for the whole system. I am not sure whether I should write something like $mgh = mgr + 1/2 m v^2 +1/2 I \omega^2 + 1/2 M u^2$ or not? Maybe I didn't completely understood the concept of rolling without slipping.

EDIT: Fixed a typo: using m instead of M in 1/2 M u^2

#### Chestermiller

Mentor
I would start by looking at the kinematics. Let $V_s(-\mathbf{i_x})$ be the velocity of the slide (i.e., in the negative x direction), and let $\mathbf{v_r}$ be the velocity of the center of mass of the ball relative to the slide. The objective is to get an expression for the velocity of the center of mass of the ball relative to a stationary observer. What is the normal component of $\mathbf{v_r}$ equal to?

#### titansarus

I would start by looking at the kinematics. Let $V_s(-\mathbf{i_x})$ be the velocity of the slide (i.e., in the negative x direction), and let $\mathbf{v_r}$ be the velocity of the center of mass of the ball relative to the slide. The objective is to get an expression for the velocity of the center of mass of the ball relative to a stationary observer. What is the normal component of $\mathbf{v_r}$ equal to?
Do you mean Normal component relative to the slide? (perpendicular to slide). I can't completely understand how to write this terms because I don't actually know any angle here. the only thing I can write is that the v of COM of ball relative to a stationary observer is $\vec{V_s} + \vec{V_r}$

#### Chestermiller

Mentor
Do you mean Normal component relative to the slide? (perpendicular to slide). I can't completely understand how to write this terms because I don't actually know any angle here. the only thing I can write is that the v of COM of ball relative to a stationary observer is $\vec{V_s} + \vec{V_r}$
Well, the normal component of the relative velocity is zero, so the relative velocity of the center is locally tangent to the slide. Let y=y(x) be the shape of the slide, and let $V_r$ represent the magnitude of the tangential relative velocity. In terms of dy/dx, and $V_r$, what are the Cartesian components of the relative velocity?

To solve this, we're going to use conservation of horizontal linear momentum and conservation of energy. To do this, we will focus only on the final velocities when the ball reaches the end of the slide. What is the vertical component of relative velocity of the center of mass of the ball at the very end of the slide?

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#### titansarus

Well, the normal component of the relative velocity is zero, so the relative velocity of the center is locally tangent to the slide. Let y=y(x) be the shape of the slide, and let $V_r$ represent the magnitude of the tangential relative velocity. In terms of dy/dx, and $V_r$, what are the Cartesian components of the relative velocity?

To solve this, we're going to use conservation of horizontal linear momentum and conservation of energy. To do this, we will focus only on the final velocities when the ball reaches the end of the slide. What is the vertical component of relative velocity of the center of mass of the ball at the very end of the slide?
At the end of slid, the vertical component will $V_r$ at that point. there will be no vertical component I think. And the speed relative to ground will be $V_r - V_s$. So shoud I write $(Initial Momentum)~~~~ 0 + 0 =(Final Momentum)~~~~ m (V_r - V_s) + M (-V_s)$ and $mgh = mgr + 1/2~ m (V_r - V_s)^2 +1/2 I \omega^2 ~ + 1/2~ M (V_s)^2$ where $\omega = R (V_r - V_s)$. Is this right? I think there must be something to do with frictions but if I write energy conservation for the whole system, the frictions will not matter because they are internal forces. So maybe some part of my equations are wrong.

#### Chestermiller

Mentor
At the end of slid, the vertical component will $V_r$ at that point. there will be no vertical component I think. And the speed relative to ground will be $V_r - V_s$. So shoud I write $(Initial Momentum)~~~~ 0 + 0 =(Final Momentum)~~~~ m (V_r - V_s) + M (-V_s)$ and $mgh = mgr + 1/2~ m (V_r - V_s)^2 +1/2 I \omega^2 ~ + 1/2~ M (V_s)^2$ where $\omega = R (V_r - V_s)$. Is this right? I think there must be something to do with frictions but if I write energy conservation for the whole system, the frictions will not matter because they are internal forces. So maybe some part of my equations are wrong.
The momentum balance looks OK, but not the energy balance. I don't think the mgr should be in there, and $\omega=V_r/R$; that is, the rate of rotation of the ball is determined only by the relative velocity of the ball with respect to the slide.

#### titansarus

The momentum balance looks OK, but not the energy balance. I don't think the mgr should be in there, and $\omega=V_r/R$; that is, the rate of rotation of the ball is determined only by the relative velocity of the ball with respect to the slide.
I think mgr is because the COM will not contact the ground. $h$ is measured from the ground to the COM. and when the ball hits the ground, COM will be $r$ meters above the ground. About $\omega$, I know it is actually $\omega r = v$ and the mistake was a typo. But can you explain why I should use $V_r$ (velocity relative to slide) and not velocity relative to ground?

Thanks.

#### Chestermiller

Mentor
I think mgr is because the COM will not contact the ground. $h$ is measured from the ground to the COM. and when the ball hits the ground, COM will be $r$ meters above the ground. About $\omega$, I know it is actually $\omega r = v$ and the mistake was a typo. But can you explain why I should use $V_r$ (velocity relative to slide) and not velocity relative to ground?

Thanks.
I agree with your mgr. From the frame of reference of the COM of the ball (i.e., as reckoned by an observer moving along with the COM of the ball), the surface of the ball in contact with the incline is moving tangentially with velocity $-\mathbf{V_r}$, so its annular velocity must be $V_r/R$. This is why they gave you the hint in the problem statement.

• titansarus

#### titansarus

I agree with your mgr. From the frame of reference of the COM of the ball (i.e., as reckoned by an observer moving along with the COM of the ball), the surface of the ball in contact with the incline is moving tangentially with velocity $-\mathbf{V_r}$, so its annular velocity must be $V_r/R$. This is why they gave you the hint in the problem statement.
Finally, It means that if I solve this two equations in terms of $V_R$ and $V_S$ I will get the answers wanted by the problem, ($u = V_s$ , $v = V_r - V_s$. Is this right? The only think odd is that in the actual question I had with numerical values, there where values for $\mu_k,\mu_s$. They didn't appear in the equations. Is this OK?

#### Chestermiller

Mentor
Finally, It means that if I solve this two equations in terms of $V_R$ and $V_S$ I will get the answers wanted by the problem, ($u = V_s$ , $v = V_r - V_s$. Is this right? The only think odd is that in the actual question I had with numerical values, there where values for $\mu_k,\mu_s$. They didn't appear in the equations. Is this OK?
I guess you can check to ascertain that the tangential force on the ball is less than the normal force times the coefficient of static friction (at least at the end).

Incidentally, you did very nicely on this problem.

#### haruspex

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Finally, It means that if I solve this two equations in terms of $V_R$ and $V_S$ I will get the answers wanted by the problem, ($u = V_s$ , $v = V_r - V_s$. Is this right? The only think odd is that in the actual question I had with numerical values, there where values for $\mu_k,\mu_s$. They didn't appear in the equations. Is this OK?
If it is going to slip, it will slip at the maximum slope. (Specifically, for rolling, $\tan(\theta_{max})=\mu(1+\frac {mr^2}I)=\frac 75\mu$. Proof: exercise.)
Since we do not know the max slope, we just have to trust the statement that it is rolling contact. That makes the given coefficients irrelevant.

Edit: typo, or maybe blunder...
$\tan(\theta_{max})=\mu(1+\frac {mr^2}I)=\frac 72\mu$.

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• Chestermiller

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