How Is Chemical Potential Energy Converted in a Skateboarding Stunt?

[flyingpig]

Homework Statement

A skateboarder with his board can be modeled as a particle of mass 79 kg, located at his center of mass (which we will study in a later chapter). As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point circle a). The half-pipe is one half of a cylinder of radius 6.90 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 6.40 m.

(b) Immediately after passing point circle b, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.870 m above the concrete (point circle c). Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 6.03 m. His body is horizontal when he passes point circle d, the far lip of the half-pipe. As he passes through point circle d, the speed of the skateboarder is 4.66 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy in the skateboarder-Earth system when he stood up at point circle b?

Homework Equations

Conservation of Energy

The actual answer is 571J

The Attempt at a Solution

$$\Delta$$KE + $$\Delta$$PE + $$\Delta$$CE = 0

m[v²/2 + g$$\Delta$$h] = $$\Delta$$CE

79(11.2²/2 + 9.8(0.5 - 6.4)] = -$$\Delta$$CE

And it isn't 571J...

What in the world is "Chemical Potential Energy"?

Also I don't know what is going on with the tex, but my delta is in lowercase for some reason, so it is supposed to be a triangle

EDIT: I am lagging really bad, so my post may not turn out very pretty

 

[tiny-tim]

hi flyingpig!

(have a delta: ∆ )

you've used KE instead of ∆KE

find ∆KE by finding the new v by using conservation of angular momentum (and why does that work? )

"Chemical Potential Energy" is is all the energy used up, including that wasted on producing heat and sound

but it doesn't matter … all they're asking is what is the difference in mechanical energy

 

[flyingpig]

hi flyingpig!

(have a delta: ∆ )

you've used KE instead of ∆KE

find ∆KE by finding the new v by using conservation of angular momentum (and why does that work? )

"Chemical Potential Energy" is is all the energy used up, including that wasted on producing heat and sound

but it doesn't matter … all they're asking is what is the difference in mechanical energy

But our class haven't learned angular momentum yet...we are suppose to tackle this problem with only energy and I used ∆KE, energy initially was 0

 

[tiny-tim]

hmm … imagine his arms were really long, and he was holding onto the centre of the circle …

when he reaches the bottom, he pulls his arms in, to pull himself up, and his angular speed will increase in exactly the same way as when a spinning skater pulls his arms in (kinetic energy is lost because the skater has to do work to pull his arms in) …

this is conservation of angular momentum …

i don't think you can do the problem without it

EDIT: actually, you could do it by calculating the work done by the reaction force instead, but that would involve an integration, and you'd need to know the centripetal acceleration formula, v²/r​

 

[flyingpig]

hmm … imagine his arms were really long, and he was holding onto the centre of the circle …

when he reaches the bottom, he pulls his arms in, to pull himself up, and his angular speed will increase in exactly the same way as when a spinning skater pulls his arms in (kinetic energy is lost because the skater has to do work to pull his arms in) …

this is conservation of angular momentum …

i don't think you can do the problem without it

EDIT: actually, you could do it by calculating the work done by the reaction force instead, but that would involve an integration, and you'd need to know the centripetal acceleration formula, v²/r​

Okay I know what centripetal acceleration is and that formula. But can you give me some more hints...?I don't know how centripetal acceleration is related to energy or where you getting at

 

[tiny-tim]

hi flyingpig!

centripetal acceleration is related to energy, because it will give you the reaction force, N, and therefore the work done, W …

use F = ma, with a = v²/r and with F = N - mg …

then use W = ∫ N.dh to give you the work done by the reaction force …

ooh, noooo, that's not going to work after all …

you still need conservation of angular momentum to find out what v is (as a function of h)

sorry, looks like you're stuck ​

 

[flyingpig]

hi flyingpig!

centripetal acceleration is related to energy, because it will give you the reaction force, N, and therefore the work done, W …

use F = ma, with a = v²/r and with F = N - mg …

then use W = ∫ N.dh to give you the work done by the reaction force …

ooh, noooo, that's not going to work after all …

you still need conservation of angular momentum to find out what v is (as a function of h)

sorry, looks like you're stuck ​

I don't think we will event ouch on angular momentum two weeks after seeing we are still doing linear momentum (1d)

Is it possible to just give me the answer...? Because I've been to khanacademy and got an idea of what angular momentum is, but there is nothing involved in energy. It is just torque

 

[tiny-tim]

conservation of angular momentum means that vr is constant, where v is (tangential) speed and r is the radius of curvature (in this case, the distance from the centre of the circle to the centre of mass) …

so when the radius decreases, the speed increases

 

[flyingpig]

conservation of angular momentum means that vr is constant, where v is (tangential) speed and r is the radius of curvature (in this case, the distance from the centre of the circle to the centre of mass) …

so when the radius decreases, the speed increases

What does that have to do with energy? We haven't touched on center of mass yet. But why can't mv^2/r dot the arc of the path work?

 

[tiny-tim]

What does that have to do with energy?

it gives you v, from which you can calculate the energy

We haven't touched on center of mass yet.

but "center of mass" is in the question!

But why can't mv^2/r dot the arc of the path work?

that would be zero (the two vectors are perpendicular) …

it has to be dot the sudden vertical change in the path

 

[sungod]

Could it be that by raising himself up, he's given himself some potential energy? Had he never raised himself up, he'd only have the potential energy due to his starting position as the initial energy and his energy at the lowest point would be that of just a particle with that specific potential energy. But, by raising himself up, his chemical energy (due to muscles etc) has added an increase in potential energy, thus having more energy at the lowest point than just a particle would have done.

 

[flyingpig]

it gives you v, from which you can calculate the energy

How do I find this constant? It was 0 initially

 

[tiny-tim]

hi sungod!

yes, flyingpig has already included ∆PE (= mgh) in his original equation …

mgh happens to equal the work done (∫ N dh) by N, the vertical reaction force, if the path is flat

but the path is curved, so there's some vertical acceleration, which changes N, which changes the work done to more than mgh

How do I find this constant? It was 0 initially

constant?

v is the two values of the speed immediately before he stands up and immediately after he stands up

 

[flyingpig]

hi sungod!

yes, flyingpig has already included ∆PE (= mgh) in his original equation …

mgh happens to equal the work done (∫ N dh) by N, the vertical reaction force, if the path is flat

but the path is curved, so there's some vertical acceleration, which changes N, which changes the work done to more than mgh


constant?

v is the two values of the speed immediately before he stands up and immediately after he stands up

0

Oh okay, I think I sort of get it now

V(0) = 11.2m/s
R = 6.4m

V(?) = ?
R = 6.03m

(11.2)(6.4) = x(6.03)

v(?) = 11.88m/s

∆mgh + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

-57.82 + 619.9288 = 562.1088J

Really close now...

 

[flyingpig]

It's due in like two days...please?

 

[tiny-tim]

V(0) = 11.2m/s
R = 6.4m

V(?) = ?
R = 6.03m

(11.2)(6.4) = x(6.03)

v(?) = 11.88m/s

∆mgh + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

-57.82 + 619.9288 = 562.1088J

(in the exam, you'll need to explain it better than that! )

Yes, this looks ok , except I don't see how you got the 57.82.

 

[flyingpig]

(in the exam, you'll need to explain it better than that! )

Yes, this looks ok , except I don't see how you got the 57.82.

∆mg(0.5 - 6.4) + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

= 3950J...

I am lost again...please help. it's due tomorrow!

 

[tiny-tim]

∆mg(0.5 - 6.4) …

oh i see …

no, i think the 0.5 is already subtracted: 6.4 = 6.9 - 0.5

hmm … I've looked at the question again, to see what the second part was, the one with D, and on second reading it looks as if the given speed at D is actually an integral part of finding the chemical energy at the bottom …

it's a weirdly long question, and perhaps you're meant to ignore almost everything in it, and just subtract the KE at A (which is zero anyway) from the KE at D (since the PE at A and D are the same)

 

[flyingpig]

oh i see …

no, i think the 0.5 is already subtracted: 6.4 = 6.9 - 0.5

hmm … I've looked at the question again, to see what the second part was, the one with D, and on second reading it looks as if the given speed at D is actually an integral part of finding the chemical energy at the bottom …

it's a weirdly long question, and perhaps you're meant to ignore almost everything in it, and just subtract the KE at A (which is zero anyway) from the KE at D (since the PE at A and D are the same)

I have 4.66^2 * 0.5 * 79 = 857J...

 

[sungod]

Quick, back-of-an-envelope kind of thing (I'm taking my zero of energy at the bottom of the original path):
Energy at A = 79*9.81*6.4 = 4954.88
Energy at B = m*(v_b)^2/2 = 4954.88 -> (v_b)^2 = 125.44
Energy at C = m*(v_b)^2 + mg(0.37) = 5246.68
Energy at D = m*(v_d)^2 + mg(6.4) = 5817.69

Energy(D) - Energy(C) = 571

(I'm taking my zero of energy as the bottom of the original path)

 

[sungod]

Sorry, I forgot to write down the factor of 1/2 for E(C) and E(D). It's just a typo. The calculation includes the 1/2.

 

[flyingpig]

Sorry, I forgot to write down the factor of 1/2 for E(C) and E(D). It's just a typo. The calculation includes the 1/2.

How did you get 125.4J??

Can you use tex or use these to help me understand?

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º ≈ w ≠ ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ⁻³
£ ¢ ¥ € ½ ½ ⅓ ⅔ ¼ ¾ ⅜ ⇌ ➔ ⬅ Å⬆⬇⬌ Ⅽ ⟫ ( ⌘ Å ▼κ ᴺ
ε ω θ ρ Ω µ Ω Δ ∑ π α τ βγδΣΥηλξφϕρσψΔ⌘Å ♢∠⊥
— | ( — ⎍ ⎎ ⌠ ⌡ ·······κΧ∟ ™
≈ ≃ ≈ ≠ ∠ — ▽ Ο
ε ω c ρ Ω µ Ω Δ ∑ π α β γ δ Σ η λ ξ φ ϕ ρ σ ψ ∂ τ
Rʀ Lʟ rᵣ

•

 

[flyingpig]

Quick, back-of-an-envelope kind of thing (I'm taking my zero of energy at the bottom of the original path):
Energy at A = 79*9.81*6.4 = 4954.88
Energy at B = m*(v_b)^2/2 = 4954.88 -> (v_b)^2 = 125.44
Energy at C = m*(v_b)^2 + mg(0.37) = 5246.68
Energy at D = m*(v_d)^2 + mg(6.4) = 5817.69

Energy(D) - Energy(C) = 571

(I'm taking my zero of energy as the bottom of the original path)

Wait I thuoght at point c, the speed changes

 

[flyingpig]

wait never mind