What is the Kinetic Energy at Point A?

[muna580]

I need help with this problem. I don't really understand what they are saying and what they are aksing me. Please can someone help me. I am guessing the whole reason I am not understandthing this problem is because of that whole "circular" concept being added to this work/energy problem.

http://img225.imageshack.us/img225/9647/untitled1jx4.jpg

 

[PhanthomJay]

I need help with this problem. I don't really understand what they are saying and what they are aksing me. Please can someone help me. I am guessing the whole reason I am not understandthing this problem is because of that whole "circular" concept being added to this work/energy problem.

http://img225.imageshack.us/img225/9647/untitled1jx4.jpg

When[/URL] there is no friction or other forces that may do work, except gravity, total mechanical energy (potential and kineti) is conserved. Write the conservation of energy equation at the beginning and at the top of the curve. You will still need to find V_top-of-curve. Apply the centripetal force equation to do so, noting that as per the problem statement, only gravity acts at this point(i.e., there is no normal force that might hold the coaster to the tracks).

Then in part 2, apply the energy conservation principle again to solve for h'.

 

[muna580]

Lets lable C to be the starting point of the rollercoster.

I hope I am doing this right

Ec = mgh + (1/2)mv^2
Ea = (2/3)mgh + Ka

What is Ka? I am having a hard time sovling for Ka. I have no idea what Ka is

Ec-Ea = total mechanical engery of the concervitive force.

Please help

 

[PhanthomJay]

Lets lable C to be the starting point of the rollercoster.

I hope I am doing this right

Ec = mgh + (1/2)mv^2
Ea = (2/3)mgh + Ka

What is Ka? I am having a hard time sovling for Ka. I have no idea what Ka is

Ec-Ea = total mechanical engery of the concervitive force.

Please help

Kinetic Energy at a is (1/2)(m)(v_a^2). But you don't know what v_a is yet. That's when you look at the centripetal force equation at a...F = mv_a^2/r, where F = mg (the force of gravity alone). Solve for v_a, then v_0 from the first set of equations. BTW, it's Ec = Ea )Initial mechanical energy = final mechanical energy when only conservative forces act.

 

[muna580]

Kinetic Energy at a is (1/2)(m)(v_a^2). But you don't know what v_a is yet. That's when you look at the centripetal force equation at a...F = mv_a^2/r, where F = mg (the force of gravity alone). Solve for v_a, then v_0 from the first set of equations. BTW, it's Ec = Ea )Initial mechanical energy = final mechanical energy when only conservative forces act.

I am still confused. Look, I made this picture, to show what I understand so far. I am a little confused when it comes to circles, that is why I am not understanding the kenetic engery at point A. Can you please expalin to me step by step, how to get kenetic engery at point A.

http://img86.imageshack.us/img86/9632/untitled1ut2.jpg

 

[PhanthomJay]

I am still confused. Look, I made this picture, to show what I understand so far. I am a little confused when it comes to circles, that is why I am not understanding the kenetic engery at point A. Can you please expalin to me step by step, how to get kenetic engery at point A.

http://img86.imageshack.us/img86/9632/untitled1ut2.jpg

Before[/URL] you can get the Kinetic Energy at point A, which is
$$KE = 1/2mv_A^2$$
you first need to find $$v_A$$.

At the top of the circle , draw a free body diagram of the coaster when it is at that point. There will be the weight of the coaster acting vertically down equal to $$mg$$, and a Normal force, N, acting on the coaster from the tracks, which might be up, down, or nothing at all. The problem statement implies that N is 0, since it states that only gravity can be counted on to supply the force necessary to keep the coaster from flying off the tracks. So therefore, only gravity is considered acting, and you apply the centripetal force equation to detrmine the speed v_A.. The centripetal force equation for motion in a circle is
$$F_{net} = mv^2/R$$, where R is the radius of the circle, v is the speed of the object tangent to the circle, and F_{net} always points toward the center of the circle (it is a centripetal or 'center-seeking' force. Please memorize this equation. Now since we have established that F_{net} = mg, and R = 25, solve for
ther unknown v_A (the mass term cancels out). Then go back and do to the Conservation of Energy thing.