Block sliding on a semicircular track with friction

[zuzelle]

Homework Statement

A block of mass m rests at point A on a circular track of radius r with coefficient of friction u. It is then released with no initial speed. Find the dependance of velocity on angle, v(theta).

Relevant Equations

Potential Energy = mgh
Kinetic Energy = mv^2/2
Work done by a force = F*S
Centripetal force = mv^2/R

Since the friction is constantly changing because of both angle change and velocity change, I took a very small angle dθ, so that while the block is covering it, its velocity would change by dv, its height by rdθ*cosθ and the friction force wouldn't change much, because N isn't changing either.

Then, I wrote the equation for centripetal force:
N - mgsin(θ) =

FBD:

Next, I thought of writing conservation of energy for this small displacement:
Potential energy + Kinetic energy = Work by friction on this small displacement + Kinetic Energy after.

So I got this:

And I am not sure if this is right, because I end up getting an equation with differentials left, and it doesn't make sense. Could someone help, please?

 

[erobz]

Share a FBD of the block at some angle $\theta$.

 

[zuzelle]

Share a FBD of the block at some angle $\theta$.

Okay

 

[erobz]

Okay

Maybe dont work with the differentials in Work-Energy right away. Write Work-Energy in the standard way and differentiate w.r.t. $\theta$. So you will have your normal terms plus an integral ( representing the work done by the friction force up to some angle $\theta$ ) in the first step. Then that result will be differentiated w.r.t $\theta$ to get the first order-nonlinear ODE.

I think it can be transformed into a linear ODE from there with a standard substition technique and solved with integration factor technique.

 

[zuzelle]

Excuse me for asking again, like this?

 

[kuruman]

Start with $~N=mg\sin\theta +m\omega^2r.~$ Using $\omega$ is preferable because linear variables ($v$) and angular variables ($\theta$) are not mixed. The force of friction is
$f_k=\mu(mg\sin\theta +m\omega^2r)$ and the Newton's second law in the tangential direction of motion is
$ma_t=mr\dfrac{d\omega}{dt}=mg\cos\theta-\mu(mg\sin\theta +m\omega^2r).$
Following @erobz's suggestion

I think it can be transformed into a linear ODE from there with a standard substition technique and solved with integration factor technique.

$\dfrac{d\omega}{dt}=\dfrac{d\omega}{d\theta}\dfrac{d\theta}{dt}=\omega\dfrac{d\omega}{d\theta}$ Now you see the reason for ditching variable $v$. The differential equation you have to solve is $$m\omega\dfrac{d\omega}{d\theta}=mg\cos\theta-\mu(mg\sin\theta +m\omega^2r).$$ It should give you $\omega(\theta)$ from which getting $v(\theta)$ is trivial since $r$ is constant.

(Edited to fix typo in equation. See posts #10 and #11)

 

[erobz]

Excuse me for asking again, like this?

Your equation on the left becomes:

$$ 0 = -mgr \sin \theta + \frac{1}{2}m v^2 + \mu m r \int \left( \frac{v^2}{r} + g \sin \theta \right) d\theta $$

Now differentiate both sides with respect to $\theta$.

EDIT: Going from Newtons Laws obviously works too (also more direct) as @kuruman showed.

 

[erobz]

Also, so we can quote your equations that follow, please take a moment to learn the math formatting from LaTeX Guide

 

[zuzelle]

Start with $~N=mg\sin\theta +m\omega^2r.~$ Using $\omega$ is preferable because linear variables ($v$) and angular variables ($\theta$) are not mixed. The force of friction is
$f_k=\mu(mg\sin\theta +m\omega^2r)$ and the Newton's second law in the tangential direction of motion is
$ma_t=mr\dfrac{d\omega}{dt}=mg\sin\theta-\mu(mg\sin\theta +m\omega^2r).$
Following @erobz's suggestion

$\dfrac{d\omega}{dt}=\dfrac{d\omega}{d\theta}\dfrac{d\theta}{dt}=\omega\dfrac{d\omega}{d\theta}$ Now you see the reason for ditching variable $v$. The differential equation you have to solve is $$m\omega\dfrac{d\omega}{d\theta}=mg\sin\theta-\mu(mg\sin\theta +m\omega^2r).$$ It should give you $\omega(\theta)$ from which getting $v(\theta)$ is trivial since $r$ is constant.

Your equation on the left becomes:

$$ 0 = -mgr \sin \theta + \frac{1}{2}m v^2 + \mu m r \int \left( \frac{v^2}{r} + g \sin \theta \right) d\theta $$

Now differentiate both sides with respect to $\theta$.

EDIT: Going from Newtons Laws obviously works too as @kuruman showed.

That makes sense, thank you

 

[TSny]

Newton's second law in the tangential direction of motion is
$ma_t=mr\dfrac{d\omega}{dt}=mg\sin\theta-\mu(mg\sin\theta +m\omega^2r).$

Check the first term on the right side. I don't think $\sin \theta$ is the correct trig function here. For example, $a_t$ should equal $g$ at the starting point where $\theta = 0$ and $\omega = 0$.

 

[kuruman]

Check the first term on the right side. I don't think $\sin \theta$ is the correct trig function here. For example, $a_t$ should equal $g$ at the starting point where $\theta = 0$ and $\omega = 0$.

Yes. One cannot have the same trig function of $\theta$ for $N$ and $a_t$ which are orthogonal. Nice catch. Typo fixed - thanks.