# Homework Help: Newtonian mechanics of a car

1. Aug 7, 2006

### Kolahal Bhattacharya

A car is driven on a large revolving platform which rotates with constant angular speed w.At t=0, a driver leaves the origin and follows a line painted radially outward on the platformwith a constant speed v0.The total weight of the car is W,and the co-eff. of friction between the car and the stage is mu.
1.Find the acceleration of the car at t>0 using polar co-ordinates.
2.Find the time in whih the car just starts to slide.
Find the direction of friction force w.r.t. the instantaneous position vector R just before the car starts to skid.

It's a big difficult problem.I followede that the path will be a spiral startig from the origin.But, cannot confidently draw the free body diagram including the polar co-ordinates.So, I could not get the equation.

2. Aug 8, 2006

### Andrew Mason

Does the platform rotate at constant angular speed after the car starts moving radially outward? I will assume it does. (It is a more interesting problem if it doesn't).

The (centripetal) acceleration of the car is $F_c(t) = -m\omega^2\vec{r}(t) = -m\omega^2v_0t\hat r$

This centripetal force has to be supplied by static friction between the wheels and the platform, so $\mu_s mg = F_c = m\omega^2v_0t$

Can you do the rest?

AM

3. Aug 13, 2006

### George Jones

Staff Emeritus
But the accleration of the car is not centripetal.

What equations give the motion of the car in polar coodinates?

4. Aug 13, 2006

### Andrew Mason

You are quite right. At any point, its velocity is the vector sum of its radial and tangential velocity. Because the tangential speed increases with increasing radius which increases linearly with time, there is a tangential acceleration as well as centripetal acceleration. The total acceleration is the vector sum of the two, which has to equal the frictional force to avoid slipping.

AM

5. Aug 13, 2006

### balakrishnan_v

The net acceleration is the coriolis acceleration and the centrepetal acceleration.
So
So the radial part is -rw^2
angular part is 2wv0

So the net acceleration is
sqrt(r^2.w^4+4w^2.v0^2)=mu.g
which gives
r=v0.t=sqrt(mu^2g^2-4w^2v0^2)/w^2

So

t=sqrt(mu^2.g^2-4w^2v0^2)/(v0.w^2)

It is clear that mu.g>2w.v0
Otherwise the car will start sliding from the start itself

Last edited: Aug 13, 2006