Maximizing Turn Radius for Sliding Passenger

[swasfy]

Homework Statement

You are driving with a friend who is sitting to your right on the passenger side of the front seat of your car. You would like to be closer to your friend so you decide to use physics to achieve your goal by making a quick turn.
If the coefficient of static friction between your friend and the car seat is 0.550 and you keep driving at a constant speed of 19m/s, what is the maximum radius (in meters) you could make your turn and still have your friend slide your way?

Homework Equations

So far: v= 19.0 m/s
a = 0
\mus= 0.55
How do I approach this problem?

The Attempt at a Solution

 

[dacruick]

Well you know first off that you are going to be using centripetal acceleration to try and get your friend closer. So you need an equation that includes speed acceleration and radius for that. You also know that you need the centripetal acceleration of your friend to match the counter-acceleration( I feel like I can't call it deceleration because they just cancel out) of the friction. So now you need an equation for friction. Once you have those two equations, backed up with some understanding, you can use them fairly easily to get your radius.

The key concept here is that the larger your radius, the less your acceleration. Imagine if the radius was infinite, that means at any point you would be traveling in a straight line, and the acceleration would be 0. Or now imagine that you turn very quickly, your acceleration is going to be large. So what you are doing is trying to find the radius at which the force of friction matches the centripetal acceleration. let me know if you need anything else

dacruick

 

[swasfy]

I can understand I need to find centripetal acceleration (by the way, is that the same as radial acceleration?) so I use a= v²/r, right? But how can I use this equation if my speed is constant? Also for friction, it's F_s= \musN where N is Newtons.
Are these the equations I need to use? Do I need to equate the two? I'm still baffled

Like this?
\murg= v² where g is 9.8m/s²? No, yes, maybe?

 

[dacruick]

yes, let me tell you, and yes. So a=v²/r will give you an acceleration. your Force = static friction x normal force. Once you get a force of friction in Newtons, you will need to find a way to compare that to an acceleration. So you will have an acceleration caused by your speed at a variable radius. and you will also have a deceleration governed by your friend's weight and the static friction constant between him/her and the seat.

 

[dacruick]

by the way, what you are trying to do is find the radius that equates those two accelerations. But i encourage you to explain the problem to me. This is one of those concepts that you really learn about when you try to convey it to others. The basis of this problem is intuitive, even if you can't put all of the pieces together yet. So try and tell me the steps you need to take to solve the problem. Even if you start by paraphrasing me.

 

[Barakn]

Technically the v in your equation is speed and not a velocity vector, and the equation is written for constant speed. The only other thing you're missing is the recognition that F=m*a.

 

[swasfy]

I got the answer right!
Ok this is how I approached it:
I know f= ma where a is acceleration, which is v^2/r
I know static friction is f=(mu)N where N is m*g
therefore: (mu)m*g=m*v^2/r
m's cancel out leaving: (mu)r*g = v^2

 

[swasfy]

Thank you guys so much for helping me out!