Ball rolling up a ramp with friction -- Find max height (Unsolved)

[srmico]

Hello,

1. Homework Statement
A spherical continuous ball is sliding with a constant velocity v along a frictionless lane. Thereafter it enters an inclined surface (the angle between the surface and the horizontal plane is α) with the coefficient of friction µ between the ball and the surface.

Homework Equations

Find the maximal height it may reach if it is known that before full stop the ball was rolling.

The Attempt at a Solution

I solved it using energies but i get an extra dependence on velocity when the ball starts rolling, and my professor told me to repeat the exercice using torque, where the solution is nicer.
I divided the motion in two parts:
1-First the ball is sliding, when it starts going up the ramp the frictional force causes it to start rotating and lose velocity until v(t1)=wR (rolling condition).
2-From rolling until it stops at v(t2)=0

1: I use v(t1)=v(0)+at => (wR)=v(0)+at, but a depends on t as well, so I don't know how to continue.

 

[Doc Al]

1: I use v(t1)^2=v(0)^2+at => (wR)^2=v(0)+at, but a depends on t as well, so I don't know how to continue.

Your equation does not seem dimensionally correct. (You have velocity squared terms added to velocity terms--that won't work.)

Apply Newton's 2nd law to translation and rotation.

 

[srmico]

Your equation does not seem dimensionally correct. (You have velocity squared terms added to velocity terms--that won't work.)

Apply Newton's 2nd law to translation and rotation.

Sorry it was a typo, i edited it.

 

[Doc Al]

Sorry it was a typo, i edited it.

They still show a mix of velocity and velocity² terms mixing together.

 

[srmico]

They still show a mix of velocity and velocity² terms mixing together.

Omg I am mixing up equations. Now should be correct. Anyway I tried doing torque=Friction*Radius=Inertia*alpha. Solving for alpha and substituting at equation w=w0+alpha*t for w=v1/r (rolling condition) i get t1=(2*v1)/(5*g*cos(z)*µ), t1 is time it takes to start rolling. v1 velocity when it starts rolling.

 

[srmico]

I still don't know how to solve the problem. Some help would be much apreciated.

 

[Doc Al]

Anyway I tried doing torque=Friction*Radius=Inertia*alpha.

Good. Now you're on the right track. This gives you an expression for alpha.

Now find an expression for the translational acceleration. (What forces act on the ball?)

Then, by combining the two, you can set up a kinematic equation to find out when the ball begins rolling without slipping.

 

[srmico]

Good. Now you're on the right track. This gives you an expression for alpha.

Now find an expression for the translational acceleration. (What forces act on the ball?)

Then, by combining the two, you can set up a kinematic equation to find out when the ball begins rolling without slipping.

I did Fg*sin(z)+Ffriction=ma, solved for a, substituting at x1=x0+vt+1/2at^2, then substituting the t for the t1 that i found and i should get the displacement for the first part?.

For the second part, when the ball is rotating I am not sure what to do. I guess now the torque changes sign, but same magnitud? Then find alpha again, this time i want to find the time for when w is 0,again, use the same displacement equation and the total displacement times sin(z) should give me the final height right?

 

[Doc Al]

I did Fg*sin(z)+Ffriction=ma, solved for a, substituting at x1=x0+vt+1/2at^2, then substituting the t for the t1 that i found and i should get the displacement for the first part?.

It's not clear to me how you solved for the time. You need both a and alpha to solve for t1. Remember: the angular velocity increases from zero while the translational velocity decreases from the initial speed.

But once you get t1, you can use that kinematic equation to get the distance traveled.

For the second part, when the ball is rotating I am not sure what to do. I guess now the torque changes sign, but same magnitud? Then find alpha again, this time i want to find the time for when w is 0,again, use the same displacement equation and the total displacement times sin(z) should give me the final height right?

Once it starts rolling without slipping, things change. For one thing, there's no more kinetic friction. But you can solve for the translational acceleration.

You can also use conservation laws.

 

[srmico]

This is what I did so far, could you tell me if it is correct?
First I will consider the motion from $t_{0}$ until the ball starts rolling at $t_{1}$. We have that:
\begin{gather*}
\tau = I\alpha \\
\tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r
\end{gather*}
Solving for $\alpha$ we get
\begin{equation}
\alpha = \frac{mg\cos(\theta)\mu_{d}\cdot r}{I}
\end{equation}
From Newton second law we also have
\begin{gather*}
\sum F = ma \\
F_{g}+F_{f}= mg\sin(\theta) + mg\cos(\theta)\mu_{d} = mg(\sin(\theta) + \cos(\theta)\mu_{d}) =ma
\end{gather*}
Solving for a
\begin{equation}
a= g(\sin(\theta) + \cos(\theta)\mu_{d})
\end{equation}
Now using the kinematics equations
\begin{gather*}
\omega_{f}-\omega_{i}=\alpha t \\
v_{f}-v_{i}=a t
\end{gather*}
Solving t and $v_{f}$ for $v_{i}=v, \omega_{i}=0, \omega_{f}=\frac{v_{1}}{r}$ (rolling condition):
\begin{gather}
t_{1} = \frac{v_{1}}{r\alpha} \\
v_{1} = v+at_{1}
\end{gather}
The displacement is
\begin{gather}
\Delta s_{1} = \frac{v_{1}^{2}-v^{2}}{2a}
\end{gather}
$\Delta s_{1}$ is a HUGE fraction.

Is it correct up to here?

 

[Doc Al]

Looks good but I would be careful with signs when you get to the kinematics step. I recommend that you let $a$ and $\alpha$ represent the magnitudes of the accelerations. I would set up the condition for rolling without slipping like so: $\alpha r t = v - a t$

But you're doing great!

 

[dean barry]

Dont you need the portion of the drag force which causes rotational torque and the remaining portion that turns linear velocity into heat ( and reduces the linear velocity) ?
This way you can calculate the point in time when rotational surface velocity = linear velocity ( rolling without slipping )

 

[Doc Al]

Dont you need the portion of the drag force which causes rotational torque and the remaining portion that turns linear velocity into heat ( and reduces the linear velocity) ?

The kinetic friction has been included in the force analysis. That's all that's needed for this part of the problem. (We are using dynamics, not energy methods here.)

 

[srmico]

Alright, it took me a while (I decided to learn some Latex) but I think I have the answer, or something close. Here is the entire Latex document as I will send it to my professor. The entire mark of the semester depends on this exercice, so feel free to comment And many thanks to Dol Al for the help!

First I will consider the motion from $t_{0}$ until the ball starts rolling at $t_{1}$. We have that:
\begin{gather*}
\tau = I\alpha \\
\tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r
\end{gather*}
Solving for $\alpha$ we get
\begin{equation}
\alpha = \frac{mg\cos(\theta)\mu_{d}\cdot r}{I}
\end{equation}
From Newton second law we also have
\begin{gather*}
\sum F = ma \\
F_{g}+F_{f}= mg\sin(\theta) + mg\cos(\theta)\mu_{d} = mg(\sin(\theta) + \cos(\theta)\mu_{d}) =ma
\end{gather*}
Solving for a
\begin{equation}
a= g(\sin(\theta) + \cos(\theta)\mu_{d})
\end{equation}
Now using the kinematics equations
\begin{gather*}
\omega_{f}-\omega_{i}=\alpha t \\
v_{f}-v_{i}=a t
\end{gather*}
Solving t and $v_{f}$ for $v_{i}=v, \omega_{i}=0, \omega_{f}=\frac{v_{1}}{r}$ (rolling condition):
\begin{gather}
t_{1} = \frac{v_{1}}{r\alpha} \\
v_{1} = v+at_{1} = v(\frac{\sin(\theta)+\frac{3}{5}\cos(\theta)\mu_{d}}{\sin(\theta)+\cos(\theta)\mu_{d}})
\end{gather}
The displacement for time $t_{1}$ is
\begin{gather}
\Delta s_{1} = \frac{v_{1}^{2}-v^{2}}{2a}
\end{gather}
Substituting (5) with $v_{1}=(4),a=(2)$
\begin{gather}
s_{1} = \frac{-2v^{2}\cos(\theta)\mu_{d}(4\cos(\theta)\mu_{d}+5\sin(\theta))}{25g(\cos(\theta)^{3}\mu_{d}^{3}+3\sin(\theta)\cos(\theta)^{2}\mu_{d}^{2}-3\cos(\theta)^{3}\mu_{d}-\sin(\theta)\cos(\theta)^{2}+3\cos(\theta)\mu_{d}+\sin(\theta))}
\end{gather}
And that's our displacement until the ball starts rolling.\\ \\
Now for the displacement until it stops, we consider Newton's second law
\begin{gather*}
\sum F = ma \\
F_{g}-F_{f}= mg\sin(\theta) - mg\cos(\theta)\mu_{s} = mg(\sin(\theta) - \cos(\theta)\mu_{s}) =ma_{1}
\end{gather*}
Solving for $a_{1}$
\begin{equation}
a_{1}= g(\sin(\theta) - \cos(\theta)\mu_{s})
\end{equation}
Now using:
\begin{gather*}
\Delta s_{2} = \frac{v_{2}^{2}-v_{1}^{2}}{2a_{1}}
\end{gather*}
Solving $\Delta s_{2}$ for $v_{2}=0$:
\begin{gather*}
\Delta s_{2} = \frac{-(v(\sin(\theta)+\frac{3}{5}\cos(\theta)\mu_{d}))^{2}}{2g(\sin(\theta)+\cos(\theta)\mu_{d})^{2}(\sin(\theta)-\cos(\theta)\mu_{s})}
\end{gather*}
The total height is
\begin{gather*}
h = (s1+\Delta s2)sin(\theta)
\end{gather*}

 

[Doc Al]

I'll try to take a look at the details of your first part when I get some time today. But I wanted to point out an error here (after the ball rolls without slipping):

And that's our displacement until the ball starts rolling.\\ \\
Now for the displacement until it stops, we consider Newton's second law
\begin{gather*}
\sum F = ma \\
F_{g}-F_{f}= mg\sin(\theta) - mg\cos(\theta)\mu_{s} = mg(\sin(\theta) - \cos(\theta)\mu_{s}) =ma_{1}
\end{gather*}

Careful! Realize that once the ball rolls without slipping, static friction acts. And that value is whatever it needs to be to prevent slipping. It is not simply $\mu N$. (That would be the maximum value, but that's not relevant here.)

 

[srmico]

I'll try to take a look at the details of your first part when I get some time today. But I wanted to point out an error here (after the ball rolls without slipping):
Careful! Realize that once the ball rolls without slipping, static friction acts. And that value is whatever it needs to be to prevent slipping. It is not simply $\mu N$. (That would be the maximum value, but that's not relevant here.)

I see what you mean, but I think it still has to be constant so it shouldn't be too important. I can't believe this was 1 problem of 12 in the exam. Not that it's so hard, but the computations are so lengthy. Anyway many thanks, if you notice something wrong let me know :)

Ps: I did most of the calculations with Maple, so if the equations are right the results should be correct.

 

[Doc Al]

I see what you mean, but I think it still has to be constant so it shouldn't be too important.

Yes, the static friction is constant. But it's not equal to $\mu mg \cos\theta$.

 

[Doc Al]

Alright, it took me a while (I decided to learn some Latex) but I think I have the answer, or something close. Here is the entire Latex document as I will send it to my professor. The entire mark of the semester depends on this exercice, so feel free to comment And many thanks to Dol Al for the help!

First I will consider the motion from $t_{0}$ until the ball starts rolling at $t_{1}$. We have that:
\begin{gather*}
\tau = I\alpha \\
\tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r
\end{gather*}
Solving for $\alpha$ we get
\begin{equation}
\alpha = \frac{mg\cos(\theta)\mu_{d}\cdot r}{I}
\end{equation}
From Newton second law we also have
\begin{gather*}
\sum F = ma \\
F_{g}+F_{f}= mg\sin(\theta) + mg\cos(\theta)\mu_{d} = mg(\sin(\theta) + \cos(\theta)\mu_{d}) =ma
\end{gather*}
Solving for a
\begin{equation}
a= g(\sin(\theta) + \cos(\theta)\mu_{d})
\end{equation}

All this looks OK.

Now using the kinematics equations
\begin{gather*}
\omega_{f}-\omega_{i}=\alpha t \\
v_{f}-v_{i}=a t
\end{gather*}

Again, careful with signs. ($a$ is opposite in direction to $v$.)

Solving t and $v_{f}$ for $v_{i}=v, \omega_{i}=0, \omega_{f}=\frac{v_{1}}{r}$ (rolling condition):
\begin{gather}
t_{1} = \frac{v_{1}}{r\alpha} \\
v_{1} = v+at_{1} = v(\frac{\sin(\theta)+\frac{3}{5}\cos(\theta)\mu_{d}}{\sin(\theta)+\cos(\theta)\mu_{d}})
\end{gather}

Not sure how you got this expression for $v_1$. Do a sanity check: What happens if friction is removed?

 

[srmico]

All this looks OK.Again, careful with signs. ($a$ is opposite in direction to $v$.)Not sure how you got this expression for $v_1$. Do a sanity check: What happens if friction is removed?

The expression for $v_1$ comes from $\omega_{f}-\omega_{i}=\alpha t$ for $\omega_{i}=0$ since it's not rolling first and $\omega_{f}=\frac{v_{1}}{r}$ rolling condition.

$v_1$ looks good to me since it has the same units as v. Do you see a mistake?

 

[Doc Al]

The expression for $v_1$ comes from $\omega_{f}-\omega_{i}=\alpha t$ for $\omega_{i}=0$ since it's not rolling first and $\omega_{f}=\frac{v_{1}}{r}$ rolling condition.

I have no problem with that starting point, but I don't see how you got your final expression for $v_1$.

$v_1$ looks good to me since it has the same units as v. Do you see a mistake?

Well, it certainly has the correct units. ;) Did you do the sanity check when $\mu \rightarrow 0$?

 

[haruspex]

The expression for $v_1$ comes from $\omega_{f}-\omega_{i}=\alpha t$ for $\omega_{i}=0$ since it's not rolling first and $\omega_{f}=\frac{v_{1}}{r}$ rolling condition.

$v_1$ looks good to me since it has the same units as v. Do you see a mistake?

Here's a different sanity check: $\theta=0$. You can solve that case quite quickly using conservation of angular momentum about a point on the ground. Friction has no moment about that point. $mvr=mv_1r+\frac 25mr^2\omega_1$.

 

[srmico]

I have no problem with that starting point, but I don't see how you got your final expression for $v_1$.Well, it certainly has the correct units. ;) Did you do the sanity check when $\mu \rightarrow 0$?

So substituting (3) on (4) we get $v_{1}-a\frac{v_{1}}{r\alpha}=v$ ==> $v_{1}(1-\frac{a}{r\alpha})=v$ ==> $v_{1}=\frac{v}{(1-\frac{a}{r\alpha})}$ And plugging the numbers I got that result. There seems to be a mistake tho. I don't know why but I would expect some g dependence for #v_{1}##

Alright since it's getting quite late (Im in germany) I will recheck all my computations tomorrow and I will make sure I get all the signs right :) Thanks for the answers! I really need to get this right!

 

[srmico]

I can't do it. Every time I try I find more mistakes. One of the equations I had was $v_{1}=\frac{v}{1-\frac{a}{r\alpha}}$. But $a=r\alpha$ is another condition for rolling, meaning that I have $v_{1}=\frac{v}{0}$. I have checked every single sign, the formulae and everything. I must be missing something, but I just don't see it.

 

[Doc Al]

One of the equations I had was $v_{1}=\frac{v}{1-\frac{a}{r\alpha}}$.

Where did that come from?

But $a=r\alpha$ is another condition for rolling

Careful! Until the ball begins rolling, that condition does not apply.

I have checked every single sign, the formulae and everything.

If you show everything step by step, someone will be able to spot your error.

I still think you may be messing up your kinematic equations. When you use $v_f = v_i + at$, realize that $a$ opposes the velocity. I would write it (as I did above) as $v_f = v_i - at$, where $a$ is the magnitude of the acceleration.

 

[dean barry]

Could you tell me why The final PE does not match the original KE ?

 

[Doc Al]

Could you tell me why The final PE does not match the original KE ?

Mechanical energy is not conserved. (Kinetic friction acts until the ball begins rolling without slipping.)

 

[dean barry]

what proportion of the friction force is attributed to linear deceleration (ie converted to heat) ?

 

[Doc Al]

what proportion of the friction force is attributed to linear deceleration (ie converted to heat) ?

The entire friction force contributes to the linear deceleration (via Newton's 2nd law). The simplest way to calculate how much mechanical energy is converted to internal energy ("heat") is to calculate the final energy and compare.

 

[srmico]

As I said in my first post, I'm not allowed to solve the exercice with energy. It has to be using torque. Sorry I haven't post anything, but the truth is that i have exams in 10 days, and I won't have time to find a final answer.

 

[Doc Al]

As I said in my first post, I'm not allowed to solve the exercice with energy. It has to be using torque.

As discussed, the way to solve the problem (at least the first part of the motion) is using dynamics (force and torque). Mechanical energy is not conserved.

Sorry I haven't post anything, but the truth is that i have exams in 10 days, and I won't have time to find a final answer.

This problem doesn't have a "neat" and simple solution. Study for your exams! That is more important than any particular problem. (Unless this one is on the test. Just kidding...I hope!)

 

[srmico]

I ended up solving it this way, if anyone is interested:
From the conservation of energy we have that
\begin{equation}
\frac{1}{2}mv^{2}=mgh+\Delta E
\label{eq:1}
\end{equation}
where $\Delta E$ is the energy difference due to friction force.
\begin{equation*}
\Delta E=E_{f}\text{(energy lost due to friction)}-E_{r}\text{(rotational energy)}
\end{equation*}
To calculate $E_{f}$ we have
\begin{equation}
F_{f}=ma=-\mu mg\cos(\theta)
\label{eq:2}
\end{equation}
\begin{equation}
E_{f}=-\mu mg\cos(\theta)d
\label{eq:3}
\end{equation}
Now from \ref{eq:2} we find
\begin{equation*}
a=-\mu g\cos(\theta)
\end{equation*}
And integrating with respect to time we find the velocity
\begin{equation*}
v=-\mu g\cos(\theta)t+v_{0}
\end{equation*}
To be able to find d, we need the time where the rolling condition is fulfilled i.e.
\begin{equation}
w(t_0)=\frac{v(t_0)}{r}
\label{eq:4}
\end{equation}
We are missing an expression for $\omega$, but we can get one from
\begin{equation*}
\tau = I\alpha = F_{f}r= \mu mg\cos(\theta)r
\end{equation*}
Solving fro $\alpha$ and using $I=\frac{2}{5}mr^{2}$
\begin{equation*}
\alpha = \frac{5}{2}\frac{\mu g \cos(\theta)}{r}
\end{equation*}
Integrating with respect to time we find
\begin{equation*}
\omega = \frac{5}{2}\frac{\mu g \cos(\theta)}{r}t
\end{equation*}
Now that we have an equation for $\omega$ and an equation for $v$ we can substitute in \ref{eq:4} to find time $t_{0}$ at which the rolling condition is met. Solving for $t_{0}$ we find
\begin{equation*}
t_{0} = \frac{2V_{0}}{7\mu g \cos(\theta)}
\end{equation*}
Since we have an expression for $v(t)$, we can integrate from $t=0$ to $t=t_{0}$ to get the distance traveled in that interval
\begin{equation*}
d=\int_{0}^{\frac{2V_{0}}{7\mu g \cos(\theta)}}(-\mu g\cos(\theta)t+v_{0}) dt= \frac{12V_{0}^{2}}{49\mu g\cos(\theta)}
\end{equation*}
Substituting $d$ at \ref{eq:3} we finally get
\begin{equation*}
E_{f}=\frac{-12}{49}mV_{0}^{2}
\end{equation*}
Now we need an expression for $E_{r}$ to see how much energy was stored as rotational energy. For that we have
\begin{equation*}
E_{r}=\frac{I\omega(t_{0})^{2}}{2}=\frac{5}{49}mV_{0}^{2}
\end{equation*}
Finally, substituting $E_{f}$ and $E_{r}$ in \ref{eq:1}
\begin{equation*}
\frac{1}{2}mv^{2}=mgh+(E_{f}-E_{r})=mgh+\frac{7}{49}mV_{0}^{2}
\end{equation*}
And solving for $h$
\begin{equation*}
h=\frac{5V_{0}^{2}}{14g}
\end{equation*}

 

[Doc Al]

A few comments.

From the conservation of energy we have that
$$\frac{1}{2}mv^{2}=mgh+\Delta E$$
where $\Delta E$ is the energy difference due to friction force.
$$\Delta E=E_{f}\text{(energy lost due to friction)}-E_{r}\text{(rotational energy)}$$

To apply conservation of energy you'd need something like: KE_i = PE_f + KE_f + rotKE_f + Energy lost.

To calculate $E_{f}$ we have

$$F_{f}=ma=-\mu mg\cos(\theta)$$

$$E_{f}=-\mu mg\cos(\theta)d$$

You have calculated the friction force times the distance the ball moved, but that is not the work done by friction, as the ball rolls as well as slides. (It's the pseudowork or center of mass work done by friction, but that's something different.)

Now from 2\ref{eq:2} we find

​

$$a=-\mu g\cos(\theta)$$

That would be the acceleration if friction were the only force. What about gravity?

 

[srmico]

I also think it's wrong, but that's what my teacher assistant told me to do. I asked him about the gravity part, and he said that it's inside the "mgh" (I know it's wrong). As I said, this is from last semester, so I will not spend any more time trying to solve it, because I don't have it (I have exams about electricity and magnetism next week).
Thanks a lot for all the answers :D

 

[Doc Al]

I asked him about the gravity part, and he said that it's inside the "mgh" (I know it's wrong).

When doing energy calculations, the work done by gravity is included in the mgh. But not for calculating the acceleration.

Good luck on your exams!