How Does Friction Affect Motion on a Rotating Car Platform?

[Patta1667]

Homework Statement

I'm not given any signs of a correct answer in the book, so could I get a check here?

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 platform with constant speed v. The total weight of the car is W, and the coefficient of friction is $$\mu$$ between the car and platform.

a) Find the acceleration of the car as a function of time using polar coordinates.

b) Find the time at which the car starts to slide.

c) Find the direction of the friction force with respect to the instantaneous position vector r just before the car starts to slide.

Homework Equations

$$F_{friction} <= \mu W$$
$$a = (r'' - r \theta' ^2) \hat r + (r \theta'' + 2r'\theta') \hat \theta$$

The Attempt at a Solution

a) r' = v, $$\theta'$$ = w, and r'' = $$\theta ''$$ = 0, so $$a = (-v w^2)t \hat r + 2vw \hat \theta$$.

b) In order for the car to follow the correct path (constant velocities in the radial and normal directions), the radial and normal accelerations must be zero. This is due to the frictional force counteracting these accelerations, and when the car starts to slide is the point at which $$m *|a_{net}| = f_{max}$$. This happens when:

$$m|a| = m\sqrt{v^2w^4t^2 + 4v^2w^2} = \mu W = \mu mg \implies t = \sqrt{\frac{u^2g^2 - 4v^2w^2}{v^2w^4}}$$

c) My figuring was that the angle between r (vector) = |r|r (unit vector) and the frictional force is the same as the angle between the r unit vector and the friction. The friction is in the opposite direction as the net acceleration force, and when the car starts to slide, $$|f| = \mu W$$. If I am using the dot product wrong here, please correct me:

$$\hat r * f = |\hat r||f| \cos \theta \implies \cos \theta = \frac{\hat r * f}{|\hat r||f|} = \frac{mvw^2}{\mu W} = \frac{vw^2}{\mu g} \implies \theta = \cos^{-1} \left( \frac{vw^2}{\mu g} \right)$$

This angle, if r is directed down the positive x-axis, should be between r and the down-right frictional force vector, correct? (the net acceleration would be pointing to the upper-left since the radial component is left and the normal component is up, so the frictional force would be pointing opposite that).

I'm not experienced with using accelerations in polar coordinates and finding angles such as in c), so I would greatly appreciate any tips or corrections. Thanks!

PS: In b), when finding |a|, is that way valid in polar vectors also? eg. $$|v| = \sqrt{v_r^2 + v_{\theta}^2}$$

 

[member 553083]

a) Yes, your answer for a) is correct.b) Yes, the way you found t is correct. You can also use the magnitude of the acceleration (which is what you did): |a| = \sqrt{(vw^2)^2 + (2vw)^2} = \sqrt{v^2w^4 + 4v^2w^2} = \sqrt{\mu^2 W^2}. Thus, t = \sqrt{\frac{\mu^2 W^2 - 4v^2w^2}{v^2w^4}}.c) Yes, your answer is correct. The friction force will always act in the opposite direction of the net acceleration force.