Block sliding on vertical track with friction (Solved problem)

[kuruman]

Homework Statement

A small block of mass $m$ is sliding on the inside of a vertical circular track of radius R. The coefficient of kinetic friction between the block and the track is μ. The block starts at the 9 o'clock position (point A) with speed $v_0$ and reaches the 3 o'clock position (point B) along two separate paths: (a) clockwise (top semicircle) or (b) counterclockwise (bottom semicircle). Assume that initial speed is high enough to keep the block in contact ith the track at all times. Gravity acts conventionally from top (12 o'clock) to bottom (6 o'clock).

Find the final speed at point B of the mass along the to paths.

Relevant Equations

$F_{net} = ma.$

This problem is a reformulation and solution of a problem posted earlier that did not go very far. A streamlined solution is posted here for future reference.

The original post had an error which has now been fixed and the solution is, once more streamlined. See posts #4 and #5 for the historical background.

Solution

Shown on the right are two FBDs for the two paths.
From the top FBD,
$N+mg\cos\theta=m\omega^2 R\implies N=m\omega^2 R-mg\cos\theta$
From the bottom FBD,
$N-mg\cos\theta=m\omega^2 R\implies N=m\omega^2 R+mg\cos\theta$
We combine the two into a single equation and write
$N=m\omega^2 R\mp mg\cos\theta.$
The force of friction for the two cases is
$f_k=\mu N=\mu m(\omega^2 R\mp g\cos\theta).$
The top/bottom sign corresponds to the top/bottom FBD. Before writing the tangential acceleration, we note that, in each case, $~-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$ Since the particle starts at A and ends at B in each case, $\sin\theta$ is negative in quadrants II (top FBD) and quadrant III (bottom FBD). The positive direction in each case is the direction of the linear velocity. Thus, the linear acceleration is
Top FBD:$~a_{\text{top}}=-\dfrac{f_k}{m}+g\sin\theta$
Bottom FBD:$~a_{\text{bot}}=-\dfrac{f_k}{m}-g\sin\theta$
The tangential acceleration for two cases combined is $$a_t=-\mu\omega^2 R \pm g(\sin\theta+ \mu\cos\theta).$$ Now $$a_t= R\frac{d\omega}{dt}=R\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega R\frac{d\omega}{d\theta}.$$This results in the differential equation $$\omega \frac{d\omega}{d\theta}=-\mu\omega^2 \pm \frac{g}{R}(\sin\theta+\mu\cos\theta)$$ which we have to solve in order to answer the question. In the absence of gravity ($g=0$), the equation becomes $~\dfrac{d\omega}{d\theta}=-\mu\omega~$ which has solution $\omega = Ce^{-\mu~\theta}.$ This motivates trying a solution $$\omega(\theta)=f(\theta)e^{-\mu~\theta}.$$ The left-hand side is $$LHS=\omega \frac{d\omega}{d\theta}=fe^{-\mu~\theta}\left(\frac{df}{d\theta}e^{-\mu~\theta}-\mu fe^{-\mu~\theta}\right)=f\frac{df}{d\theta}e^{-2\mu~\theta}-\mu\omega^2.$$ Setting this equal to the RHS and canceling what cancels yields a separable equation, $$f\frac{df}{d\theta}=\pm \frac{g}{R}(\sin\theta+ \mu\cos\theta)e^{2\mu~\theta} \implies f~df=\pm g(\sin\theta+\mu\cos\theta)e^{2\mu~\theta}d\theta.$$ Using Wolfram Alpha for the integral on the RHS, we obtain $$
\begin{align} & \frac{1}{2}\left(f^2-f_0^2\right)=\pm\frac{g}{R}\frac{\left[(2\mu^2-1)\cos\theta+3\mu\sin\theta \right]}{[(4\mu^2+1)}{e^{2\mu \theta}} \nonumber \\
& f=\left\{f_0^2 \pm\frac{2g}{R}\frac{\left[(2\mu^2-1)\cos\theta+3\mu\sin\theta \right]}{(4\mu^2+1)}{e^{2\mu \theta}} \right\}^{1/2} \nonumber \\
& \implies \omega=\left\{f_0^2 \pm\frac{2g}{R}\frac{\left[(2\mu^2-1)\cos\theta+3\mu\sin\theta \right]}{(4\mu^2+1)}{e^{2\mu \theta}} \right\}^{1/2}e^{-\mu~\theta} \nonumber \\
\end{align}$$ The initial condition is $\omega(-\pi/2)=v_0/R$ and can be used to find $f_0$. $$ \begin{align}
& \frac{v_0}{R}=\left [ f_0^2 \pm\frac{2g}{R}\frac{(-3)\mu}{(4\mu^2+1)}{e^{-\mu \pi}} \right]^{1/2}e^{\mu \pi/2}. \nonumber \\
& f_0^2=\left[ \frac{v_0^2}{R^2}\pm \frac{6\mu g}{(4\mu^2+1)R}\right]e^{-\mu \pi}. \nonumber \\
\end{align}$$ Finally, $$\omega= \left \{\left[ \frac{v_0^2}{R^2}\pm \frac{6\mu g}{(4\mu^2+1)R}\right]e^{-\mu( \pi+2\theta)} \pm\frac{2g}{R}\frac{\left[(2\mu^2-1)\cos\theta+3\mu\sin\theta \right]}{(4\mu^2+1)} \right\}^{1/2}.$$ At last we are in a position to find the speed at point B. $$\begin{align}
&\omega_B=\omega(\pi/2)= \left \{\left[ \frac{v_0^2}{R^2}\pm \frac{6\mu g}{(4\mu^2+1)R}\right]e^{-2\mu\pi} \pm\frac{6\mu g}{(4\mu^2+1)R} \right\}^{1/2}. \nonumber \\
& \omega_B= \left[\frac{v_0^2}{R^2}e^{-2\mu\pi}\pm\frac{6\mu g}{(4\mu^2+1)R}(1+e^{-2\mu\pi}) \right]^{1/2}\nonumber \\
& \implies v_B=\left[ v_0^2e^{-2\mu\pi}\pm \frac{6\mu gR}{(4\mu^2+1)}(1+e^{-2\mu\pi})\right]^{1/2} . \nonumber \\
\end{align}$$

 

[Juanda]

One quick question. I find it odd the first expression you use lacks a $-$ sign. It might just be related to how you defined the positive directions of your coordinates but I wanted to confirm it because in the original thread, I already had doubts about the sign convention (post #35) and I still couldn't fully solve them.

You wrote:

From the top FBD,
$N+mg\cos\theta=m\omega^2 R$

However, if positive $r$ grows outwards and positive $\theta$ grows counterclockwise, then it'd have a minus sign. Also, I'm having trouble to see how $N$ and $mg\cos\theta$ could be positive as well. They point inwards making $r$ smaller so it feels like they'd be negative. Similar to how in a xy coordinate system, a left-pointing force is negative when it accelerates the body in the negative direction.
This page shows the derivation for the formulas. It's unfortunate it uses $\phi$ instead of $\theta$ and also it's got some missing equations due to formatting errors but the main idea is still there.

Can you add how you defined $+r$ and $+\theta$? I assume $+r$ is pointing outwards and I'm guessing $+\theta$ is pointing clockwise instead of counterclockwise which might explain why you didn't get that $-$ sign.

 

[kuruman]

One quick question. I find it odd the first expression you use lacks a $-$ sign. It might just be related to how you defined the positive directions of your coordinates but I wanted to confirm it because in the original thread, I already had doubts about the sign convention (post #35) and I still couldn't fully solve them.

You wroteowever, if positive $r$ grows outwards and positive $\theta$ grows counterclockwise, then it'd have a minus sign. Also, I'm having trouble to see how $N$ and $mg\cos\theta$ could be positive as well. They point inwards making $r$ smaller so it feels like they'd be negative. Similar to how in a xy coordinate system, a left-pointing force is negative when it accelerates the body in the negative direction.

Strictly speaking you are correct. If you look at the top FBD, the radial components of $\vec N$ and $m\vec g$ are negative and so is the centripetal acceleration. This means that the equation should have been written conventionally as $$-N-mg\cos\theta=-m\omega^2 R.$$However, if you multiply both sides of the equation by $-1$, you get all the negative signs converted to positive signs which is what I have. What matters is the relative signs between terms, not the overall sign.

This page shows the derivation for the formulas. It's unfortunate it uses $\phi$ instead of $\theta$ and also it's got some missing equations due to formatting errors but the main idea is still there.
View attachment 334232

I don't disagree with this page and there is no misfortune to using $\phi$ instead of $\theta$. You can see how these more general forms become what I have if you set $\phi=\theta$, $\dot \phi=\omega$ and $r=R=\rm{constant}$:
$F_r=-m\omega^2R$
$F_{\theta}=mR\dfrac{d\omega}{dt}.$

Can you add how you defined $+r$ and $+\theta$? I assume $+r$ is pointing outwards and I'm guessing $+\theta$ is pointing clockwise instead of counterclockwise which might explain why you didn't get that $-$ sign.

Angle $\theta$ is shown in the drawing of the two FBDs.
For the top diagram, $\theta$ increases clockwise as the mass moves from point A to point B. At A $\theta =-\frac{\pi}{2}$ and at B $\theta =+\frac{\pi}{2}.$
For the bottom diagram, $\theta$ increases counterclockwise as the mass moves from point A to point B. At A $\theta =-\frac{\pi}{2}$ and at B $\theta =+\frac{\pi}{2}.$

 

[haruspex]

Bottom FBD:$~a_{\text{bot}}=-\dfrac{f_k}{m}-g\sin\theta$
The tangential acceleration for two cases is $$a_t=-\frac{f_k}{m}\pm g\sin\theta=-\mu(\omega^2 R\mp g\cos\theta)\pm g\sin\theta=-\mu\omega^2 R \pm g(\sin\theta-\mu\cos\theta).$$

Isn’t the sign wrong on that last $\mu\cos\theta$ term?
As you have it, that frictional component is reinforcing the acceleration due to gravity. It should be opposing it even in the top case (because it comes from a reduction in the normal force, so reduces the downward friction).

 

[kuruman]

As you have it, that frictional component is reinforcing the acceleration due to gravity.

I think you're right. Let me redo it in more detail justifying the signs for the benefit of anyone who visits this thread.

See figure on the right. The motion is from A to B, clockwise for the top track and counterclockwise for the bottom track. In both cases angle $theta$ varies from $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi.$ Thus, $\sin\!\theta$ is negative in quadrants (II) and (III) and positive in quadrants (I) and (IV). The positive tangential direction is in the direction of the motion, so $f_k$ must be negative in all quadrants. Writing the 1d tangential vector as $$f_k=-\mu N=-\mu m(\omega^2 R\mp g\cos\!\theta)$$ does the job as friction always opposes the motion. Of course, if $f_k$ changes sign for some angle (top track), this means that the normal force has gone through zero at which point the block will fall off.

If we write the component of gravity as $mg\sin\!\theta$, we must be careful to ensure that its sign is positive when the mass is moving downhill and gravity aids the motion and negative when the mass is moving uphill and gravity opposes the motion. So
Quadrant (I): $~f_g=mg\sin\!\theta.$ Motion downhill, $\sin\!\theta>0$, overall sign positive.
Quadrant (II): $~f_g=mg\sin\!\theta.$ Motion uphill, $\sin\!\theta<0$, overall sign negative.
Quadrant (III): $~f_g=-mg\sin\!\theta.$ Motion downhill, $\sin\!\theta<0$, overall sign positive.
Quadrant (IV): $~f_g=-mg\sin\!\theta.$ Motion uphill, $\sin\!\theta>0$, overall sign negative.
For the two tracks the total tangential accelerations are $$\begin{align}
& a_{\text{top}}=-\mu (\omega^2 R- g\cos\!\theta)+g\sin\!\theta=-\mu \omega^2 R+g(\sin\!\theta+\mu \cos\!\theta)
\nonumber \\
& a_{\text{bot}}=-\mu (\omega^2 R+ g\cos\!\theta)-g\sin\!\theta=-\mu \omega^2 R-g(\sin\!\theta+\mu \cos\!\theta)
\nonumber \end{align}$$ Putting the two together gives the expression in post #1, $$a_{\text{t}}=-\mu \omega^2 R \pm g(\sin\!\theta+\mu \cos\!\theta).$$ The relative sign in the parentheses should be $+$ instead of $-$. I appreciate you spotting this. One good turn deserves another. At some point I will have to repost the solved problem. Ugh!