# Finding kinetic energy

1. May 29, 2015

### kaspis245

1. The problem statement, all variables and given/known data
A body is moving on a horizontal plane which gradually turns to another plane forming an angle $α$. Friction coefficient is $μ$. Find the kinetic energy of the body after the turn if in the beginning it was $K_o$.

2. Relevant equations
Newton's laws.

3. The attempt at a solution
A frictional force $F_{fr}=μmg$ is applied to the body, but since I don't know the distance over which it was applied for, I can't find a proper equation. I can also add that at the end of the turn the body will be affected by a force equal to $F_{tr}+mg⋅sinα$ which is directed in parallel to the plane.

$K_o=mgh+K$ , so if I could find an expression for $h$ the problem would be solved (I suppose the mass would cancel out).

Last edited: May 29, 2015
2. May 29, 2015

### DEvens

Could you give your answer as a function of h? That would seem to satisfy both parts of your difficulties.

3. May 29, 2015

### CWatters

Perhaps I'm missing something... It's a round body so presumably friction is causing it to roll. Rolling resistance isn't mentioned.

4. May 29, 2015

### haruspex

Yes, but it gets tricky if $\mu$ is small compared with $\tan(\alpha)$.
Kaspis, is this the whole question, word for word?

5. May 29, 2015

### kaspis245

Absolutely yes.
Rolling resistance is not given so I suppose we have to assume that the body doesn't roll.

6. May 30, 2015

### haruspex

No, I wouldn't draw that conclusion. Rolling resistance is normally very small, so neglected.
As the question stands, I would answer K0. There's no way to put a lower bound on the height change in executing the turn. We can say the height gain is at least r(1-cos(α)), but by making r sufficiently small this becomes insignificant compared to K0.
If you do make it a function of h, you need to think about whether rolling contact will be maintained. For some parameter settings, it would be lost at first then regained later. And you would need to know both coefficients of friction.

7. May 30, 2015

### ehild

You can consider the body point-like, moving along an arc of circle. Gravity G, normal force N, and friction f act on the body. Find the resultant tangential and radial force in terms of θ, the angle of the turn. You get a differential equation for v2.

8. May 30, 2015

### haruspex

How is this any different in result from just writing down lost KE = gained PE?

9. May 30, 2015

### ehild

Nothing is said about rolling or about the shape of the body, but the coefficient of friction is indicated. That suggest sliding, so friction also does work and changes the kinetic energy.

10. May 30, 2015

### haruspex

Ok, so throw in a $\mu R \alpha$ term - still no need for a differential equation. And still not really an answer since we don't know the radius of curvature.

11. May 30, 2015

### ehild

We do not know a lot of things so the problem in this form has no sense. We can interpret it in different ways.
If the body rolls, the static friction is b less or equal μ times the normal force along the path, but the mechanical energy is conserved.

If the body slides, the force of friction is μ times the normal force, and it does work.
In both cases, we need to know the normal force. As it is a curved path, the normal force depends of the speed and the radius of the curvature.
I do not understand, where to "throw in a μRα therm". It is length not work.

Last edited: May 30, 2015
12. May 30, 2015

### haruspex

true.

13. May 30, 2015

### kaspis245

While the body is moving along an arc of a circle it is affected by radial force equal to $N$, so:
$N=m\frac{v^2}{R}$
$mg⋅cosα=m\frac{v^2}{R}$
$g⋅cosα=\frac{v^2}{R}$

It is also affected by tangential force which is equal to downward directed forces:
$F_T=F_{fr}+mg⋅sinα=μmg⋅cosα+mg⋅sinα$
$a_T=g(μ⋅cosα+sinα)$

I am not really sure what to do now.

14. May 30, 2015

### ehild

That is not correct. The centripetal force is the resultant of the normal force and the radial component of gravity.
The expression for the force of friction is not right. The force of friction is μ times the normal force, which is not mgcosα now.
If you have the correct expression for the tangential force, you can write the tangential acceleration, which is dv/dt (v is the speed).

15. May 31, 2015

### kaspis245

If centripetal force is the resultant of the normal force and the radial component of gravity, so:
$mg⋅cosα-mω^2R=m\frac{v_2}{R}$

Why the normal force is not equal to $mg⋅cosα$? Does it equal to the sum of radial force and $mg⋅cosα$?

16. May 31, 2015

### ehild

What do you mean on 'radial force'?
The normal force is N. The centripetal force is mv2/R which is the same as mω2R.
The resultant of the normal force and the radial component of gravity is equal to the centripetal force: N-mgcosθ=mω2R.

17. May 31, 2015

### kaspis245

I see, so $N=mω^2R+mgcosα$, then the tangential force is equal to:

$F_T=μ(mω^2R+mgcosα)+mgsinα$

$a_T=μ(ω^2R+gcosα)+gsinα$

18. May 31, 2015

### ehild

It is almost right now, but the tangential force opposes the motion, so it needs a negative sign. And you should use some other angle, (say θ), as alpha is the angle corresponding to the final position, The tangential acceleration can be written in terms of the angular acceleration $a_T=R \dot \omega$
So the equation becomes $R\dot \omega=-μ(ω^2R+gcos(\theta)-gsin(\theta)$.
This is a differential equation for omega. Are you familiar with differential equations?

19. Jun 2, 2015

### kaspis245

A bit. So I need to express $w'$ and find it's antiderivative?

20. Jun 2, 2015

### ehild

The equation
$R\dot \omega=-μ(ω^2R+gcos(\theta))-gsin(\theta)$
is for ω(t), angular velocity in terms of the time. But you need the KE in terms of the angle. The equation does not depend explicitly on time so you can change the variable to θ. Apply chain rule and express dω/dt with dω/dθ.