Angular momentum and a summersault problem

[robax25]

Homework Statement

A jumper departs from an inclined plane with a velocity of 100km/h and no rotation. He wants to summersault while he is airborne within 2s.Based on the conservation of angular momentum, he will start tom rotate if his wheel's rotation is slowed down. His wheels are hollow cylinder with a radious of 0.4m and a mass of 10kg. The inertia of the bike and him is 200kgm².Calculate how much the wheels' rotation has to be modified in order to make a full rotation of the bike during 2s.Let's assume the angular velocity can be changed within 0.1s. which angular acceleration does that correspond to?Find out
1) Initial angular velocity of the wheel.
2)required angular momentum of the bike.
3)Modified angular velocity of the bike
4)Angular acceleration

Homework Equations

Angular momentum L=I*omega. [/B]

The Attempt at a Solution

1) omega ω= 69.44 rad/s.
2)Angular momentum L=13888 kgm²/s
3)
4)Angular acceleration =694.4 rad/s²

 

[gneill]

Did you have a particular question to ask about the problem? You haven't shown any attempt, just some answers (yours?) for certain portions of the problem.

Can you tell us what you're having problem with, your thoughts about it, and show us what you've tried?

 

[robax25]

There is no answer to my hand.I just do all calculation to my own effort.I am uncertain what should I do to calculate modified velocity?I have showed my attempt.To calculate modified velocity, I need to find out angular acceleration , as velocity goes to slow down.Then to find final angular velocity I need to use this equation ωf=ωo -∝t (minus because it is negative acceleration)

 

[haruspex]

I agree with your answer for the first part, but not the second. How did you arrive at that? What will be the rider's rotation rate for most of the 2 s?

You cannot answer part 4 until you have solved part 3. You seem to have assumed the rider will bring the wheels to a stop relative to the bike. That need not be so.

 

[robax25]

you are right. First two questions are easy to solve. However, 3rd and 4th I have confusion. I do not know how I proceed.For full rotation, he needs to decelerate. But How I get acceleration?. I did that α=ω/t so 69.6/2=-34.8 rad/s²

 

[haruspex]

So what do you get for part 2 now, and how do you get it? What you had in post #1 was wrong.

The question order is a bit odd. The logical sequence is 1-3-2-4. How far must the rider rotate, and in what time?

How I get acceleration?

As I wrote, need to answer part 3 first.

 

[robax25]

you are right. First two questions are easy to solve. However, 3rd and 4th I have confusion. I do not know how I proceed.For full rotation, he needs to decelerate. But How I get acceleration?. I did that α=ω/t so 69.6/2=-34.8 rad/s²

That is my idea for part 3,

 

[haruspex]

That is my idea for part 3,

Part 3 does not involve acceleration.
Try to answer my questions in post 6.

 

[robax25]

Time is 2s to modify wheel's rotation.ω=2*pi/T=2*3,1416/2=3.1416rad/s I did to get required angular momentum L=I*ω=200kgm²*69.44 rad/s=13888kgm²/s,Yes it is wrong. He needs to rotate 2*pi rad

 

[haruspex]

Time is 2s to modify wheel's rotation

No, 2s is the time to modify the angle (not the rotation) of the bike and rider (not the wheels).

ω=2*pi/T=2*3,1416/2=3.1416rad/s

Yes, that is the rate of rotation the bike and rider need to maintain for the 2s. That answers part 3.

L=I*ω=200kgm²*69.44 rad/s=13888kgm²/s,Yes it is wrong.

So what is the right answer to part 2?

 

[robax25]

No, it is not the right answer. The right answer for part 2, Required momentum is L=I*ω=200kgm²*3.1416rad/s=628.32 kgm²/s and For part 4, Angular acceleration= ω/T =3.1416rad/s/0.1s=31.416 rad/s²

 

[haruspex]

No, it is not the right answer. The right answer for part 2, Required momentum is L=I*ω=200kgm²*3.1416rad/s=628.32 kgm²/s and For part 4, Angular acceleration= ω/T =3.1416rad/s/0.1s=31.416 rad/s²

Good.

 

[robax25]

Calculate how much the wheels' rotation has to be modified in order to make a full rotation of the bike during 2s? I do not understand the sentence.It says that wheel's rotation has to be modified...but we do not do it...is it possible to calculate?I mean we have initial angular velocity and then, we goes slow down to do full rotation of the bike. However, in same initial angular velocity i cannot do it.I have confusion there

 

[haruspex]

Calculate how much the wheels' rotation has to be modified in order to make a full rotation of the bike during 2s? I do not understand the sentence.It says that wheel's rotation has to be modified...but we do not do it...is it possible to calculate?I mean we have initial angular velocity and then, we goes slow down to do full rotation of the bike. However, in same initial angular velocity i cannot do it.I have confusion there

You calculated the rotation the bike+rider needs to acquire. How does it do this? At take off, the wheels are rotating but the bike is not. What can the rider do to transfer wheel rotation to bike rotation?

 

[robax25]

he does breaking. I mean he decelerates.I just write according to the question

 

[haruspex]

he does breaking. I mean he decelerates.I just write according to the question

And why does that result in rotation of the bike?

 

[robax25]

because he feels negative force instantly.

 

[haruspex]

because he feels negative force instantly.

Braking is not going to decelerate the bike while he is airborne. What will it do?

 

[robax25]

you are right. The reason to have a rotation because of angular momentum is constant during flight. There is no torque acting on the bike.There is a huge force acting on it due to acceleration or inclined jump.He reduces his velocity/ acceleration because of gravity acting on the center point of bike and him.It also happens in projectile motion.

 

[haruspex]

The reason to have a rotation because of angular momentum is constant during flight.

Right. So can you answer the question now?