# Angular momentum and a skateboarder's halfpipe

• kylecarson
In summary, the skateboarder with a height of 1.5m and weight of 60 kilograms drops down a half-pipe with a radius of 4 meters using a frictionless skateboard. She starts with her center of gravity at 1/4 her height and then pushes it up to 1/2 her height at the bottom of the half-pipe. Using the equations for linear and angular momentum, we can calculate that she will catch 7.75 meters of air above the other side of the half-pipe.
kylecarson

## Homework Statement

A skate boarder with height 1.5m and weight 60 kilograms
drops down a half-pipe with a radius of 4 meters using her
frictionless skateboard. At first, she carries her center of
gravity really low on the skateboard, say 1/4 her height.
At the bottom of the half-pipe, she pushes her center of
gravity up to 1/2 her height. She then continues up the other
side of the pipe. How much air does she catch (above the other side
of the half-pipe)?

## Homework Equations

1/2mv2 + mgh = 1/2mv2 + mgh + Wapp
L = I Ω (omega)
Ω = v/r
I = m r
L = mr (v /r ) = mr (v /r )
r1v1 = r2v2

## The Attempt at a Solution

d = (√(2g (r + h/4)) ( r-h/4)/(r-h/2))^2/ 2g
d = (r + h/4) (( r-h/4)/(r-h/2))^2
d = 4m + (1.5m/4)((4m-(1.5m/4))/(4m-(1.5m/2)))^2/(2 (9.81m/s^2))

d = 4m + (1.5m/4)((3.5m/4)/(3m))^2/ (2 (9.81m/s^2)) d = 4m + (1.5m/4)((7/4)^2/(3))^2/ (2 (9.81m/s^2)) d = 4m + (1.5m/4)(49/9)/(2 (9.81m/s^2)) d = 4m + (1.5m/4)(49/19.62) d = 4m + (1.5m/4)(2.5) d = 4m + 3.75m d = 7.75m

= 2.5m

I would approach this problem by first understanding the concept of angular momentum and how it relates to the motion of the skateboarder in the half-pipe. Angular momentum is a property of a rotating object and is defined as the product of its moment of inertia (I) and angular velocity (ω). In this case, the skateboarder's motion can be described using the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Next, I would consider the conservation of angular momentum in this scenario. According to this principle, the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the skateboarder has no external torques acting on her, so her angular momentum will remain constant throughout her motion in the half-pipe.

To determine the final height of the skateboarder, I would use the equation L = Iω and the fact that ω = v/r, where v is the linear velocity and r is the radius of the half-pipe. I would also consider the change in the skateboarder's moment of inertia as she changes her center of gravity from 1/4 her height to 1/2 her height. This change in moment of inertia would result in a change in her angular velocity, which would ultimately determine the height she reaches on the other side of the half-pipe.

Based on these considerations, I would calculate the final height using a combination of equations and principles, taking into account the initial and final positions of the skateboarder's center of gravity and the conservation of angular momentum. The exact calculation would depend on the specific values given in the problem, but the final answer should be in line with the given solution of 2.5m.

## 1. What is angular momentum?

Angular momentum is a property of a rotating object, such as a skateboarder moving through a halfpipe. It is defined as the product of an object's moment of inertia and its angular velocity.

## 2. How does angular momentum affect a skateboarder in a halfpipe?

As a skateboarder moves through a halfpipe, they are constantly changing their direction and rotation. Angular momentum helps them maintain their balance and control their movements by providing stability and resistance to rotational changes.

## 3. Can a skateboarder change their angular momentum in a halfpipe?

Yes, a skateboarder can change their angular momentum in a halfpipe by altering their body position, speed, and direction of movement. This allows them to perform tricks and maneuvers in the air and on the walls of the halfpipe.

## 4. How does the shape of a halfpipe affect a skateboarder's angular momentum?

The shape of a halfpipe can greatly influence a skateboarder's angular momentum. A steeper halfpipe will provide more gravitational potential energy, allowing the skateboarder to generate more angular momentum and perform higher and more complex tricks. A shallower halfpipe may limit the skateboarder's momentum and restrict their movements.

## 5. How do friction and air resistance affect a skateboarder's angular momentum in a halfpipe?

Friction and air resistance can both impact a skateboarder's angular momentum in a halfpipe. Friction can slow down the skateboarder's rotational movements, while air resistance can make it more difficult for the skateboarder to maintain their balance and control. Skaters must account for these factors when performing in a halfpipe to maintain their angular momentum and execute their tricks properly.

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