# Particle on inclined plane whose steepness is increasing

1. Jul 8, 2015

• Missing homework template due to originally being posted in other forum.
I am trying to solve a problem using Newtonian Mechanics. I was able to solve it using Lagrangian Mechanics, and got the correct answer from the textbook, but when I use Newtonian I get the wrong answer. I want to be able to do it both ways in order to understand/appreciate the difference between Newtonian and Lagrangian.

The problem
A particle of mass m rests on a smooth plane. The plane is raised to an inclination angle θ at a constant rate α (θ=0 at t=0), causing the particle to move down the plane. Determine the motion of the particle.

My attempt at a solution
If we designate the starting position of the particle to be r0 (distance from the axis around which the plane rises), and r(t) to be the distance travelled by the particle down the ramp at a given time t, then r(t)=∫r'(t)dt from 0→t, where r'(t)=∫r''(t)dt from 0→t.
Now r''(t) is just the component of gravity directed down the ramp, so r''(t)=g*sin(αt).
∴ r'(t)=∫g*sin(αt)dt=g/α (1-cos(αt)). Then r(t)=∫r'(t)dt=g/α∫(1-cos(αt))dt=g/α (t - 1/α sin(αt))

If you do it using Lagrangian mechanics, you get r0(1-cosh(αt))-(g/2α^2)*(sin(αt)-sinh(αt)).

Any insight on logical errors I have made with my Newtonian approach would be greatly appreciated. Thanks!!

2. Jul 8, 2015

### Dazed&Confused

The equation of motion in polar coordinates in the radial direction is $F_r = m \ddot{r} -mr\dot{\theta}^2.$

Last edited: Jul 8, 2015
3. Jul 8, 2015

Hi Dazed&Confused, thank you for your reply. Can you explain to me how you came up with that and why my assumption was incorrect? I thought F_r was just the component of gravity directed down the plane at a given time, so mg*sin(αt)

4. Jul 8, 2015

### Nathanael

Your assumption was that ar=r'' which is not true. As Dazed&Confused pointed out, there is another term in the radial acceleration (what people call centripetal acceleration).

In cartesian coordinates, the unit vectors are constant in time, so you can just differentiate twice to get ax=x''
But with polar coordinates, you are using coordinates which vary in time (if θ varies in time) and so you get additional terms.

If you look up 'acceleration in polar coordinates' then I'm sure there are many derivations of the full acceleration formula.
(Or you can find it yourself, but first you have to figure out $\frac{d}{dt}\hat r$ and $\frac{d}{dt}\hat \theta$ which are not zero unless dθ/dt is zero)

5. Jul 8, 2015

### Dazed&Confused

You can also derive this formula with the Lagrangian approach. With the Lagrangian $$\frac12m(\dot{r}^2 + r^2\alpha^2) -mgr\sin\alpha t$$ you will find the same formula with $F_r = -mg\sin\alpha t.$