Okay. There's a lot here, I know, but I really need help! I have an exam coming up. I will put in as much information as possible. Even if you don't fully solve it, can you just help me get started so I can solve it?
A mass m = 72 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 18.9 m and finally a flat straight section at the same height as the center of the loop (18.9 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)
1. What is the minimum speed the block must have at the top of the loop to make it around the loop without leaving the track?
Answer: 13.62 m/s
----My Solution----
Starting with Ki+Ui=Kf+Uf, I get the equation mg=mv^2/4
m's cancel leaving g=v^2/r. so I take g and get sqrt(gr) and get 13.62
Online program says this is correct!
2. What height above the ground must the mass begin to make it around the loop-the-loop?
Answer: 47.25 m
----My Solution----
Starting with Ki+Ui=Kf+Uf I get the equation mgh=1/2mv^2=answer+(2r)
so h=1/2v^2/g and add 2r.
Answer: 47.25 which the online program says is correct.
3. If the mass has just enoughs peed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?
---This one doesn't tell me whether the answer is right or wrong so I have no idea what to do. I've tried many different routes.---
4. If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.9 m off the ground)?
---I need to know how to do number 3 to be able to do this one.---
5. Now a spring constant k=18100 N/m is used on the final flat surface to stop the mass. How far the does the spring compress?
A mass m = 72 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 18.9 m and finally a flat straight section at the same height as the center of the loop (18.9 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)
1. What is the minimum speed the block must have at the top of the loop to make it around the loop without leaving the track?
Answer: 13.62 m/s
----My Solution----
Starting with Ki+Ui=Kf+Uf, I get the equation mg=mv^2/4
m's cancel leaving g=v^2/r. so I take g and get sqrt(gr) and get 13.62
Online program says this is correct!
2. What height above the ground must the mass begin to make it around the loop-the-loop?
Answer: 47.25 m
----My Solution----
Starting with Ki+Ui=Kf+Uf I get the equation mgh=1/2mv^2=answer+(2r)
so h=1/2v^2/g and add 2r.
Answer: 47.25 which the online program says is correct.
3. If the mass has just enoughs peed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?
---This one doesn't tell me whether the answer is right or wrong so I have no idea what to do. I've tried many different routes.---
4. If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.9 m off the ground)?
---I need to know how to do number 3 to be able to do this one.---
5. Now a spring constant k=18100 N/m is used on the final flat surface to stop the mass. How far the does the spring compress?
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3.
Energy: ½mVt² + m*g*(2*R) = ½m*Vb² → Vb² = Vt² + 4*g*R
Vb = 30.44 m/s
4.
Same reasoning as (3): Vf² = Vt² + 2*g*R → Vf = 23.6 m/s
5.
Es = KE
½kx² = ½mVf² → x² = m*Vf²/k
x = 1.49 m
Energy: ½mVt² + m*g*(2*R) = ½m*Vb² → Vb² = Vt² + 4*g*R
Vb = 30.44 m/s
4.
Same reasoning as (3): Vf² = Vt² + 2*g*R → Vf = 23.6 m/s
5.
Es = KE
½kx² = ½mVf² → x² = m*Vf²/k
x = 1.49 m