# Block sliding on vertical track with friction (Solved problem)

• kuruman
kuruman
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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.

Solution
Shown below 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 is $$a_t=-\frac{f_k}{m}\pm g\sin\theta=-\mu(\omega^2 R\mp g\cos\theta)\pm g\sin\theta=-\mu\omega^2 R \pm g(\sin\theta-\mu\cos\theta).$$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)=\mp\frac{g}{R}\frac{(2\mu^2\cos\theta-\mu\sin\theta+\cos\theta)e^{2\mu \theta}}{4\mu^2+1} \nonumber \\ & f=\left [f_0^2 \mp \frac{2g}{R} \frac{(2\mu^2\cos\theta-\mu\sin\theta+\cos\theta)e^{2\mu \theta}} {4\mu^2+1} \right]^{1/2} \nonumber \\ & \implies \omega= \left [f_0^2 \mp \frac{2g}{R} \frac{(2\mu^2\cos\theta -\mu\sin\theta+\cos\theta)e^{2\mu \theta}}{4\mu^2+1} \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 \mp\frac{2g}{R}\frac{\mu e^{-\mu \pi}}{(4\mu^2+1)} \right]^{1/2}e^{\mu \pi/2}. \nonumber \\ & f_0^2=\left[ \frac{v_0^2}{R^2}\pm \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi}. \nonumber \\ \end{align} Finally, $$\omega= \left \{\left[ \frac{v_0^2}{R^2}\pm \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi} \mp \frac{2g}{R} \frac{(2\mu^2\cos\theta -\mu\sin\theta+\cos\theta)e^{2\mu \theta}}{4\mu^2+1} \right\}^{1/2}e^{-\mu~\theta}.$$ We test this expression by setting ##\mu =0## whch is the frictionless case. \begin{align} & \omega= \left( \frac{v_0^2}{R^2} \mp \frac{2g}{R}\cos\theta\right)^{1/2} \nonumber \\ & \frac{v^2}{R^2}=\frac{v_0^2}{R^2} \mp \frac{2g}{R}\cos\theta \nonumber \\ & \frac{1}{2}mv^2=\frac{1}{2}mv_0^2 \mp mgR\cos\theta \nonumber \\ & \frac{1}{2}mv_0^2=\frac{1}{2}mv^2 \pm mgR\cos\theta. \nonumber \\ \end{align}The last equation is the familiar statement of mechanical energy conservation for the top and bottom paths with the potential energy being zero at the horizontal diameter.

At last we are in a position to find the speed at point B. \begin{align} &\omega_B= \left \{\left[ \frac{v_0^2}{R^2}\pm \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi} \mp \frac{2g}{R} \frac{( -\mu)e^{\mu \pi}}{(4\mu^2+1)} \right\}^{1/2}e^{-\mu~\pi/2} \nonumber \\ & \omega_B= \left[\frac{v_0^2}{R^2}e^{-2\mu\pi}\pm\frac{2\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{2\mu gR}{(4\mu^2+1)}(1+e^{-2\mu\pi})\right]^{1/2} . \nonumber \\ \end{align}
Shown below is a plot of ##v(\theta)=\omega(\theta)~R## for the two paths with plotting parameters as shown.

Last edited:
Juanda, anuttarasammyak, MatinSAR and 1 other person
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:
kuruman said:
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.

Juanda said:
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.
Juanda said:
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}.##

Juanda said:
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}.##

Juanda

## 1. What forces act on the block as it slides down the vertical track?

As the block slides down the vertical track, the forces acting on it include gravity (acting downward), the normal force (perpendicular to the track surface), and the frictional force (opposing the motion of the block).

## 2. How do you calculate the frictional force on the block?

The frictional force can be calculated using the equation $$f = \mu N$$, where $$\mu$$ is the coefficient of friction and $$N$$ is the normal force. On a vertical track, the normal force is influenced by the component of the gravitational force perpendicular to the track surface.

## 3. What is the role of the normal force in this problem?

The normal force is crucial because it determines the magnitude of the frictional force. It acts perpendicular to the surface of the track and is influenced by the gravitational force acting on the block and the angle of the track.

## 4. How does the angle of the track affect the motion of the block?

The angle of the track affects both the normal force and the component of gravitational force parallel to the track. As the angle increases, the parallel component of gravity increases, causing the block to accelerate more quickly, while the normal force decreases, reducing the frictional force.

## 5. How can you determine the acceleration of the block?

The acceleration of the block can be determined by applying Newton's second law in the direction of motion. This involves summing the forces parallel to the track (gravity component and friction) and dividing by the mass of the block, resulting in the equation $$a = \frac{mg \sin(\theta) - f}{m}$$, where $$m$$ is the mass of the block, $$g$$ is the acceleration due to gravity, $$\theta$$ is the angle of the track, and $$f$$ is the frictional force.

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