Finding the time when the car starts to skid (Kleppner 2.29)

  • #1

Homework Statement


This problem is taken from Kleppner's Intoduction to Mechanics and is problem 2.29.

A car is driven on a large revolving platform which rotates with constant angular speed [itex]\omega[/itex]. At time [itex]t=0[/itex] a driver leaves the origin and follows a line painted radially outward on the platform with constant speed [itex]v_0[/itex]. The total weight of the car is [itex]W[/itex], and the coefficient of friction between the car and stage is [itex]\mu[/itex].

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 skid.


Homework Equations



Acceleration in polar coordinates [itex](\dot{r}\ -r \dot{\theta}^2 )\hat{r} +(r \ddot{\theta} +2\dot{r}\dot{\theta})\hat{\theta}[/itex].

[itex] f_{MAX} =\mu W[/itex], where [itex] f [/itex] is friction.

The Attempt at a Solution



So the acceleration is [itex](-v_0t\omega^2 )\hat{r} +(2v_0\omega)\hat{\theta}[/itex].

I think the time the car begins to skid when the frictional force cannot provide the necessary acceleration, or when the acceleration is [itex] \geq \mu g[/itex]. I would then find the absolute value of the acceleration, equate it to [itex]\mu g[/itex], and solve for [itex]t[/itex]. I'm not sure if this is correct, however.
 

Answers and Replies

  • #2
verty
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I have meticulously verified that this is correct. It turns out the exponential formula ##r e^{j\theta}## is safe to use which is nice.

##x(t) = v_0te^{jwt}##
##a(t) = [- v_0 w^2 t + 2vwj] e^{jwt}##

And I agree that ##|a(t)| = \mu g## is the correct calculation to do.
 
Last edited:
  • #3
Well using that I get [tex] t = \dfrac{1}{\omega}\sqrt{\left(\dfrac{\mu^2g^2}{v_0^2\omega^2} -4\right)} [/tex].
 
  • #4
verty
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Right, you've solved it. It's a strange formula but it must be correct.
 
  • #5
Right, you've solved it. It's a strange formula but it must be correct.

The thing is part c. asks me to find the frictional force with respect to the instantaneous position vector r just before it starts to skid. I think this could give a messy result.
 
  • #6
verty
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The thing is part c. asks me to find the frictional force with respect to the instantaneous position vector r just before it starts to skid. I think this could give a messy result.

Hmm, so they want mass times acceleration at time t, which requires substituting t into a(t).

##a(t) = v_0 w [-wt \hat{r} + 2 \hat{\theta}]##

I see no value in substituting for t here. I suppose this question was about managing complexity and being confident of each step. I certainly learned something doing it. So I suppose it wasn't such a bad question after all.
 
  • #7
Hmm, so they want mass times acceleration at time t, which requires substituting t into a(t).

##a(t) = v_0 w [-wt \hat{r} + 2 \hat{\theta}]##

I see no value in substituting for t here. I suppose this question was about managing complexity and being confident of each step. I certainly learned something doing it. So I suppose it wasn't such a bad question after all.

Sorry I read the question incorrectly. It was actually to find the direction of the frictional force which would require [itex]\text{arctan}[/itex].
 
  • #8
verty
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I would just skip that part, that is too complicated. And anyway it's just the angle of the acceleration and one could always find it by substituting the values first and then finding the angle.
 
  • #9
I would just skip that part, that is too complicated. And anyway it's just the angle of the acceleration and one could always find it by substituting the values first and then finding the angle.

Ok. Thank you for your help.
 

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