Can't Get Started on Car Acceleration Problem. Please Help!

[Double A]

Can't Get Started. Please Help!

The following question I have spent the past few hours trying to figure out how to solve it. I'm not sure where to start.

Question:
A car traveling on an flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.70 m/s^2. The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data.

I have been looking through the corresponding section in the text to this question for an understanding/example solution/equations but I'm not seeing anything that helps me.

Anyone have an idea?

 

[Justin Lazear]

You know that the centripetal force that the static friction must provide is

$$F_c = m \omega^2 R$$

Do you know what $\omega^2$ is when the car slips? Would you be able to determine this (at least as a function of radius) if you knew the acceleration?

You know that the maximum value $F_c$ can take is the maximum static friction, i.e. $F_s = \mu mg$. Can you solve for $\mu$ with this information?

--J

 

[Double A]

Sorry, but I'm a little confused. Is $$F_c$$ the centripetal force? The only information I'm given is in the question above. Are you saying that I need to first do some symbolic algebra?

 

[Justin Lazear]

Get used to having to deal with letters. Because radius is not necessary to solve this problem, we don't need a numerical value for it, even though our equations involve radius. It simply means that it's going to cancel out somewhere and the final equation will not. Personally, I don't even consider the numerical values of expressions until the absolute end.

Not including the force which is providing the 1.70 m/s^2 acceleration to the car, (we already know everything we care to know about that particular one, i.e. that it accelerates the car 1.70 m/s^2 in the direction of motion), there is only the force of static friction in the system. The force of static friction is the centripetal force. However, we have two different ways of calculating what this force is. We can calculate it by noting that the car follows a circular path and is hence in circular motion, and we can calculate it directly using the static friction equation I cited below. Regardless of how we do it, there is only one force, so these two methods must yield the same answer, hence we can equate them.

--J

 

[Double A]

Ok, I have been playing around with what you have been saying but I do not think that I am still getting it.

Your telling me that:
$$F_c=F_s$$

Which is: $$m \omega^2 R = \mu_s m g$$

Solving for $$\mu_s$$: $$\mu_s = \frac{\omega^2 R}{g}$$

I'm not seeing where the radius is canceling and I am not sure what $$\omega^2$$ is or how to find it. Is there something that I am not getting or missing?

 

[Double A]

I think I have it now. When I went back and dug a little more in the text I found that $$a_c = \frac{v^2}{r} = \omega^2 r$$.

So, since $$F_c = F_s$$

Then $$m \omega^2 r = \mu_s m g$$

$$m a_c = \mu_s m g$$

Mass cancels. So, then solving for $$\mu_s$$

$$\mu_s = \frac{a_c}{g}$$

Putting in values where $$a_c = 1.70 \frac{m}{s^2}$$ and $$g = 9.8 \frac{m}{s^2}$$

$$\mu_s = \frac{1.70}{9.8} = 0.173$$

Thank you for leading me in the right direction. I was confused at first but as I did more digging in the text I found the right relationship that lead me to the solution. I appreciate the time you gave to help me.

 

[Justin Lazear]

You're on the right track, but $a_c$ is not what you think it is.

$a_c$ is the centripetal acceleration. You're going to use the relation $a_c = \omega^2 r$, but you need to find $\omega^2$.

If you have the linear acceleration, how would you calculate the angular acceleration? How would you use this angular acceleration to find $\omega^2$ given that you know the distance through which it was accelerating (i.e. do you have an equation for angular acceleration, angular velocity, and angle that doesn't include time?) ?

--J

 

[Double A]

I have the equation $$\omega_f^2 = \omega_i^2 + 2\alpha(\theta_f-\theta_i)$$

Solving for $$\alpha$$

$$\alpha = \frac{\Delta \omega^2}{2 \Delta \theta}$$

Because the car travels one quarter of a circle $$\theta = \frac{\pi}{2}$$

So then $$\alpha = \frac{\Delta \omega^2}{\pi}$$

Where do I go from here? I do not know what $$\alpha$$ and $$\omega^2$$ are.

The only other relationship I know of is that the total acceleration of the system is equivalent to the addtion of the tangential acceleration vector with the centripetal acceleration vector. Does this help me find what $$\alpha$$ is? So then I could use what I derived above to find $$\omega^2$$?

 

[Justin Lazear]

You're almost there!

You actually do know $\alpha$. You're given the linear acceleration $a$ of the car, and the linear acceleration is related to angular acceleration in circular motion by

$$\alpha r = a$$

And you're right, you don't know what $\omega^2$ is. But that's okay! Remember the equation you had,

$$m \omega^2 r = \mu_s m g$$ ?

If you had an expression for $\omega^2$, and this expression just happened to havean r in the denominator, do you think you could solve for $\mu_s$ in this problem?

--J

 

[Double A]

Ok, after some long hard thought and mounds of scratch paper this is what i finally came up with. Rather than using the equations associated with angular motion I used equations that are associated with linear motion.

Here is what I did:

$$m a_c = \mu g r$$

$$m\frac{v_f^2}{r} = \mu m g$$

$$v_f = \sqrt{\mu g r}$$

$$v_f^2 = v_i^2+2a_t(\frac{1}{4})(2\pi r)$$

$$v_f^2 = 0 + a_t \pi r$$

$$v_f = \sqrt{a_t \pi r}$$

$$\sqrt{\mu g r} = \sqrt{a_t \pi r}$$

$$\mu g r = a_t \pi r$$

$$\mu g = a_t \pi$$

$$\mu = \frac{a_t \pi}{g}$$

$$\mu = \frac{1.7 \pi}{9.8} = 0.545$$

This should be the right answer because I was able to talk with my instructor and he guided me in this direction. So, thanks for all the help hopefully you can help me on some more problems in the near future because I have a few more problems on this assignment that I'm not getting the right answer for. I'll post them up as soon as I get more time which may not be until tomorrow.