Moments of Inertia, Kinetic Energy and rotations

In summary: So the initial energy would be the sum of the potential energy at the top of the hill and the kinetic energy from the bike's initial velocity. Then, at the bottom of the hill, the final energy would be the sum of the kinetic energy from the bike's final velocity and the rotational energy of the wheel (since it will have no potential energy at the bottom). In summary, the bicycle racer's wheel comes off while going downhill at 11.0 m/s and a height of 75.0 m. To determine the wheel's speed at the foot of the hill and its total kinetic energy at the bottom, conservation of energy must be applied. The initial energy is the sum of potential energy and kinetic energy from the bike
  • #1
lsatwd
5
0

Homework Statement


A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes.

How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down? How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

Homework Equations


I think...
mgy = 1/2*m*v^2 + 1/2*I*w^2
I=mr^2
v=rw ("w" is supposed to be angular velocity)

The Attempt at a Solution



I'm totally stumped by part A. I just don't see how i can solve for its final speed without knowing the time or the degree. I'm also assuming the question is asking for its linear speed since it wants the answer in m/s.

I thought about using mgy = 1/2*m*v^2 + 1/2*I*w^2
= squareroot{4gy} = v

but this didn't work and I was thinking that this didn't work because the wheel isn't falling straight down from a height of 75 m, but rather rolling down from what would be the hypotenuse of a triangle with that height...but i understand a triangle isn't needed in this problem. (but I'm not sure)

I then tried finding the angular velocity by using the given speed of the wheel, 11 m/s (which i assumed to be equal to the linear velocity of the wheel) by using the equation v=rw ([11m/2]/.425 = w) which gave me a value of 25.88 rad/s to plug into the equation: 1/2*2.25kg*11^2 + 1/2(2.25*.425^2)25.88^2 = 136.30 J for total kinetic energy... but this was incorrect. I'm also not sure why the total kinetic energy isn't just mgy which would equal 1653.75 J; this number is a lot bigger than 136.30 and I'm just really confused :mad:
 
Physics news on Phys.org
  • #2
mgy = 1/2*m*v^2 + 1/2*I*w^2
I=mr^2
This doesn't look right. Conservation of energy means that

Initial energy = final energy

So for starters, what is the initial energy? Remember that the bike was going 11.0 m/s initially.

I then tried finding the angular velocity ... by using the equation v=rw
Yes, that is helpful to relate v and w.
 
  • #3
I was thinking that its total initial energy would be the sum of its translational and rotational kinetic energy so that is why i thought i could use that equation.

I don't know what to use if that's not it.
 
  • #4
lsatwd said:
I was thinking that its total initial energy would be the sum of its translational and rotational kinetic energy ...

In addition to that, potential energy should be part of the sum also.
 

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation.

2. How is moment of inertia related to rotational kinetic energy?

The moment of inertia of an object determines how much rotational kinetic energy it possesses. Objects with larger moments of inertia require more energy to rotate at the same speed compared to objects with smaller moments of inertia.

3. How is rotational kinetic energy different from linear kinetic energy?

Rotational kinetic energy is the energy possessed by an object due to its rotation, while linear kinetic energy is the energy of an object due to its motion in a straight line. They are calculated differently, with rotational kinetic energy being dependent on both the moment of inertia and the angular velocity of the object.

4. Can an object have both linear and rotational kinetic energy?

Yes, an object can have both linear and rotational kinetic energy at the same time. This is often the case for objects that are both moving and rotating, such as a spinning top or a rolling ball.

5. How can the moment of inertia of an object be changed?

The moment of inertia of an object can be changed by altering its mass distribution or its shape. For example, a figure skater can change their moment of inertia by bringing their arms in or extending them out while spinning, which affects their rotational speed.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
311
  • Introductory Physics Homework Help
Replies
3
Views
801
  • Introductory Physics Homework Help
Replies
7
Views
906
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
959
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
624
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
726
Back
Top