Angular momentum and a summersault problem

In summary: I have shown 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)For part 4, Angular acceleration= ω/T =3.1416rad/s/0.1s=31.416 rad/s²
  • #1
robax25
238
3

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
 

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  • #2
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?
 
  • #3
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)
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
robax25 said:
How I get acceleration?
As I wrote, need to answer part 3 first.
 
  • #7
robax25 said:
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,
 
  • #8
robax25 said:
That is my idea for part 3,
Part 3 does not involve acceleration.
Try to answer my questions in post 6.
 
  • #9
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
 
  • #10
robax25 said:
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).
robax25 said:
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.
robax25 said:
L=I*ω=200kgm²*69.44 rad/s=13888kgm²/s,Yes it is wrong.
So what is the right answer to part 2?
 
  • #11
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
 
  • #12
robax25 said:
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.
 
  • #13
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
 
Last edited:
  • #14
robax25 said:
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?
 
  • #15
he does breaking. I mean he decelerates.I just write according to the question
 
  • #16
robax25 said:
he does breaking. I mean he decelerates.I just write according to the question
And why does that result in rotation of the bike?
 
  • #17
because he feels negative force instantly.
 
  • #18
robax25 said:
because he feels negative force instantly.
Braking is not going to decelerate the bike while he is airborne. What will it do?
 
  • #19
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.
 
  • #20
robax25 said:
The reason to have a rotation because of angular momentum is constant during flight.
Right. So can you answer the question now?
 

Related to Angular momentum and a summersault problem

1. What is angular momentum?

Angular momentum is a measurement of an object's rotational motion. It is determined by the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum related to a summersault?

In a summersault, the body rotates around a central axis. This rotation creates angular momentum, which helps to maintain the motion of the summersault.

3. How does angular momentum affect the landing of a summersault?

As the body completes the summersault, it will have a certain amount of angular momentum. This momentum must be conserved, so the body will continue to rotate until it is either slowed down or stopped by an external force.

4. Can angular momentum be changed during a summersault?

Yes, the amount of angular momentum can be changed by altering the body's moment of inertia or angular velocity. This can be done by changing the body's shape or mass, or by changing the speed of the summersault.

5. How can angular momentum be calculated for a summersault?

To calculate the angular momentum of a summersault, you will need to know the body's moment of inertia and angular velocity. These values can be calculated using the body's mass, shape, and speed. The formula for angular momentum is L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.

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