# Problem about an incomplete Loop-the-loop

#### rivendell

1. The problem statement, all variables and given/known data
An automobile of mass M drives onto a loop-the-loop, as shown. (click here for diagram) The minimum speed for going completely around the loop without falling oﬀ is v0. However, the automobile drives at constant speed v, where v < v0. The coeﬃcient of friction between the auto and the track is μ. Find an equation for the angle θ where the auto starts to slip. There is no need to solve the equation.

2. Relevant equations & 3. The attempt at a solution
FBD equations (where f represents F of friction)
Radial direction --> N-Wcosθ = M(v^2/R)
Tangential direction --> -f = MR(θ") = m(dv/dt)
so I get dv/dt = -f/M (1)
and N = Wcosθ + M(v^2/R) (2)
Then I substitute (2) into (1) to get dv/dt = -u(gcosθ + v^2/R).
But I can't solve this differential equation. I'd only be able to do that if I didn't consider gravity at all in this problem. If I could solve it, then I'd be able to get an equation for θ. Should I not consider gravity to make solving the diff eq easier? If these steps so far are right and I can continue this way, how can solve the diff eq I have above? Thank you!

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#### TSny

Homework Helper
Gold Member
If the car travels at constant speed, what is the value of dv/dt?

Is friction the only force acting in the tangential direction?

Does friction on the car act in the direction of increasing θ or decreasing θ?

#### rivendell

thanks for noting that! I'll correct my writeup:
-f - Wsinθ = MR(θ") = M(dv/dt)
--> dv/dt = -f/M - gsinθ
--> N = Wcosθ + M(v^2/R)
Subst: dv/dt = (-ugcosθ + v^2/R)/M - gsinθ = 0
I guess I can't solve this eq, which is ok, as long as it's right. friction acts in direction of decreasing θ, which I think my eqs show. Is this right?

#### TSny

Homework Helper
Gold Member
--> dv/dt = -f/M - gsinθ
Still have a problem with the direction of f. Does f try to push the car down the slope or does it try to prevent the car from slipping down the slope?

--> N = Wcosθ + M(v^2/R)
Looks good.

Subst: dv/dt = (-ugcosθ + v^2/R)/M - gsinθ = 0
I don't believe you substituted correctly here. Check this.

I guess I can't solve this eq, which is ok, as long as it's right.
You could solve for θ using a trig identity to relate sinθ and cosθ. But you are not asked to do that.

#### rivendell

I thought f was just going against θ, so I had written that.
Should I have written dv/dt = f/M - gsinθ instead?
If I did that, then trying again subbing, dv/dt = u(Mgcos + Mv^2/R)/M - gsinθ = u(gcosθ + v^2/R) - gsinθ = 0. Does that work? and just if I wanted to solve for θ, would I use the sin double-angle identity (sin2θ = 2sinθcosθ)? thanks!

#### TSny

Homework Helper
Gold Member
I thought f was just going against θ, so I had written that.
Should I have written dv/dt = f/M - gsinθ instead?
Think about a car going at constant speed up a slope. No acceleration implies that the net force along the slope is zero. If gravity has a component down the slope, then what direction does friction need to act in order to make the net force zero?
If I did that, then trying again subbing, dv/dt = u(Mgcos + Mv^2/R)/M - gsinθ = u(gcosθ + v^2/R) - gsinθ = 0. Does that work?
Yes. Looks right.

and just if I wanted to solve for θ, would I use the sin double-angle identity (sin2θ = 2sinθcosθ)?
I don't see how that identity would be directly helpful. Use sin2θ +cos2θ = 1 to write cosθ in terms of sinθ. Then you can get the equation in terms of one unknown: sinθ. You should be able to manipulate it into a quadratic equation for sinθ.

#### rivendell

got it. Thank you TSny!

#### Matt Hebert

Why does net force in the tangential direction have to equal zero? I thought both the friction force and the force of the gravitational component would both oppose the direction of motion. Also, the question asks when the car starts slipping, does this mean when vT=0?

#### TSny

Homework Helper
Gold Member
Hello, Matt. Welcome to PF.

Friction does not always act opposite to the direction of motion. Sometimes, the friction force acting on an object is in the same direction as the motion of the object. A good example is a car starting from rest and accelerating along a horizontal surface. What force acting on the car causes the car to accelerate?

In the loop-the-loop problem, the car moves at constant speed along the circle. So, the car has zero acceleration in the tangential direction of the circle.

The point at which the car starts slipping is the point where the static force of friction between the tires and the loop is at maximum possible value.

#### Matt Hebert

Oh ok, i was treating this as a block problem, this makes a lot more sence, thanx!

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