Why does the speed change at the bottom of the loop-the-loop?

[Vigorous]

Homework Statement

A roller coaster of mass M is at the top of the Loop-the-loop of radius R at twice the
minimum speed possible. What force does the track exert on it? What force does it
exert when it is at the bottom of the circle? (Use conservation of energy if needed.)

Relevant Equations

ac=mv^2/R

1 At the top of the loop-the-loop, the track cannot pull the roller coaster upward so it acts along with the weight of the roller coaster towards the center
N+Mg=Ma=Mv^2/R
N=M(v^2/R-g)=M(4Rg/R-g)=3Mg
N cannot act upwards, therefore vmin=(Rg)^1/2. Since its stated in the problem that the roller coaster is moving at twice the minimum speed, its speed is 2(Rg)^1/2.
At the bottom of the circle
If we also analyze the forces in the radial direction at the bottom of the circle
N-Mg=Mv^2/R=4Mg N=5Mg
My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.

 

[Doc Al]

My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.

Careful. Don't confuse this situation with uniform circular motion, where the speed is constant.

 

[Lnewqban]

Car at top of loop has enough velocity as to keep moving horizontally (until gravity takes over and it starts describing a parabolic trajectory); therefore, it needs some force from the track to keep it describing a circular trajectory.
Car at bottom: same situation, only that the car has higher velocity and requires greater corrective force.

 

[Vigorous]

1) Thank you for the clarification, I drew the free body diagram again and considering the points above if the weight force is resolved along the tangential and radial direction it will provide tangential acceleration. But how can we determine the changes in the normal force and speed as it goes from top to bottom given the initial speed without resorting to the law of conservation of energy. I thought about writing the displacement velocity and acceleration in the radial and tangential directions but I haven't still been exposed to that material or may someone send me any resources explaining how to rewrite these vectors in terms of polar axes.

 

[Doc Al]

But how can we determine the changes in the normal force and speed as it goes from top to bottom given the initial speed without resorting to the law of conservation of energy.

What do you have against using conservation of energy?

There's a reason they only ask you to analyze forces at the top and bottom of the circular path: Those points are much easier to analyze since the acceleration is purely centripetal. And since the acceleration depends upon the speed, conservation of energy is the easy way to find the speed at the bottom.

 

[kuruman]

To add to what @Doc Al posted, this problem is no different from a pendulum bob at the end of a massless rod that is allowed to go all the way around from zero to 2π. You can write down the equation of motion, but you cannot solve it analytically to get the velocity as a function of time. That is why one resorts to the small angle approximation to describe the oscillatory motion of a simple pendulum.

Mechanical energy conservation is equivalent to doing the so called first integral which allows you to find the velocity as a function of position. It involves the standard transformation $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}.$$