Calculating Kinetic Energy After a Turn on a Horizontal Plane

[kaspis245]

Homework Statement

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 $K_o$.

Homework Equations

Newton's laws.

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).

 

[DEvens]

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

 

[CWatters]

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

 

[haruspex]

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

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

 

[kaspis245]

Kaspis, is this the whole question, word for word?

Absolutely yes.

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

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

 

[haruspex]

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

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.

 

[ehild]

Homework Statement

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 $K_o$.

Homework Equations

Newton's laws.

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).

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.

 

[haruspex]

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

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

 

[ehild]

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

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.

 

[haruspex]

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.

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.

 

[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.

 

[haruspex]

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.

true.

 

[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.

 

[ehild]

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

That is not correct. The centripetal force is the resultant of the normal force and the radial component of gravity.

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

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).

 

[kaspis245]

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

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.

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

 

[ehild]

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 $mg⋅cosα$?

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.

 

[kaspis245]

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

$F_T=μ(mω^2R+mgcosα)+mgsinα$

$a_T=μ(ω^2R+gcosα)+gsinα$

 

[ehild]

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

$F_T=μ(mω^2R+mgcosα)+mgsinα$

$a_T=μ(ω^2R+gcosα)+gsinα$

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?

 

[kaspis245]

This is a differential equation for omega. Are you familiar with differential equations?

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

 

[ehild]

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

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.

 

[kaspis245]

I am having a hard time understanding this. Is this correct:
$R(ω(ω(θ)))=-μ(ω^2R+gcos(θ))-gsin(θ)$

 

[ehild]

I do not see what you mean.
You know that ω=dθ/dt.
By the chain rule, dω/dt = (dω/dθ)(dθ/dt)= (dω/dθ) ω.
With that, the original equation transforms into
$Rω \frac{dω}{dθ} = -μ(ω^2R+gcos(θ))-gsin(θ)$, a differential equation for ω(θ).
The left side can be written as (1/2) d(ω² )dθ. You can consider the ω² = z the dependent variable instead of ω. So we arrived at the equation
$0.5Rz'= -μ(zR+gcos(θ))-gsin(θ)$

Rearranged:
$z' + 2μz =-\frac{2g}{R}\left(μ\cos(θ)+g\sin(θ)\right)$
It is a first-order linear differential equation for z. Do you know how to solve such equations?

 

[theodoros.mihos]

If radian is not given then curvation on shape may be here for not assume ping-pong on the corner.

 

[kaspis245]

It is a first-order linear differential equation for z. Do you know how to solve such equations?

Unfortunately, no. I think I've reached the limits of my knowledge. Here's my try:

$z=-\frac{2g(gcos(θ)-μsin(θ))}{R}$

 

[theodoros.mihos]

I cannot see data for $R$ for the moment. Is this a problem given?

 

[kaspis245]

I cannot see data for $R$ for the moment. Is this a problem given?

No, $R$ is not given in the problem statement. I believe it will cancel out when we'll find the solution.

 

[theodoros.mihos]

So, ignore the curving part. Take as two line paths with different conditions.

 

[kaspis245]

But the problem states that there is an angle $a$, so it must appear in my final equation. Besides, I don't think we can treat the body's trajectory as a line because it's length differs from that of an arc and that is crucial when there is a frictional force involved.

 

[theodoros.mihos]

For this reason you must work with different conditions on two paths, before and after the corner.

 

[ehild]

Unfortunately, no. I think I've reached the limits of my knowledge. Here's my try:

Do not worry, you will learn it...
By the way, it might be that I misunderstood the problem, and it was a rolling body as Haruspex said.. And the condition that 'one plane gradually turns to the other one' only means that the body does not bounce. And we will consider it a collision problem. I'll think about it. Are you sure you copied the original problem text word-by word?

 

[kaspis245]

Are you sure you copied the original problem text word-by word?

Absolutely yes. Notice that it's not mention that it's a ball. I think we were going in the right direction.

 

[ehild]

Absolutely yes. Notice that it's not mention that it's a ball. I think we were going in the right direction.

If you want to go ahead with the differential equation, see:
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html
We have theta instead of x as the independent variable and z(theta) is the function instead of y(x),
$z' + 2μz =-\frac{2g}{R}\left(μ\cos(θ)+g\sin(θ)\right)$
So p(θ)=2μ and its integral is 2μθ. That is simple.
$u(θ)=e^{\int(p(θ)dθ}=e^{2μθ}$
Now the tricky part comes: Find the integral $S=\int{u(θ)g(θ)dθ}=-\frac{2g}{R}\int{e^{2μθ}\left(μ\cos(θ)+g\sin(θ)\right)dθ}$ Integrate by parts twice.
The solution is $z(θ)=\frac{S+C}{u(θ)}$
C is the integration constant,the initial value of z, (z at θ=0).

 

[mfb]

As the radius of curvature is not given, the intended answer must be independent of it. This is not true in general - even without friction the loss to gravity will depend on R. It is true if we let R be very small and treat the mass as (non-rolling) point, however. That leads to a unique solution that depends on the given quantities only. Not sure if you can get away without integration, but it certainly makes the equations easier.

 

[ehild]

As the radius of curvature is not given, the intended answer must be independent of it. This is not true in general - even without friction the loss to gravity will depend on R. It is true if we let R be very small and treat the mass as (non-rolling) point, however. That leads to a unique solution that depends on the given quantities only. Not sure if you can get away without integration, but it certainly makes the equations easier.

So you suggest to solve the problem for the limiting case R-->0, but still assuming smooth transition between the two planes.
In case of very small radius, the transition from one plane to the other happens in a very short time. It is collision, in principle. The normal force is impulsive, and gravity can be ignored with respect to it. Friction is also impulsive, as it it proportional to the normal force. The change of kinetic energy is equal to the work of all forces. The work of the normal force is zero, The work of gravity can be ignored, so it is the work of friction alone to be counted with.
The change of KE is equal to the work of friction $dE= -μN(Rdθ)$ and N is equal to the centripetal force if mg is ignored : $N=\frac{mv^2}{R}$, so
$dE= -μ\frac{mv^2}{R}(Rdθ) = -μmv^2dθ$.
$E=1/2 mv^2$, so we can write a DE for the kinetic energy:
$\frac{dE}{dθ} = -2μE$.
It is very easy to solve and get the kinetic energy at the final angle α.
@https://www.physicsforums.com/members/kaspis245.534046/

 

[haruspex]

It is collision, in principle.

Treating it as an inelastic collision would mean that the component of velocity normal to the ramp would be lost in the impact, independently of any friction. Your analysis above does not do that, but I agree with mfb that that analysis is exactly what is wanted in answering the question.