Energy changes of a ball thrown up

  • Thread starter Thread starter SunshineCat
  • Start date Start date
  • Tags Tags
    Ball Energy
Click For Summary
SUMMARY

The discussion focuses on the energy changes of a hollow ball thrown vertically upwards, analyzing two scenarios: one where the ball is not rotating and another where it is solid. The ball's mass is 0.310 kg, radius is 0.0340 m, and it is released with a linear speed of 5.24 m/s and a rotational speed of 2.70 revolutions per second. In the first scenario, the height reached will be less than 1.40 m due to decreased total kinetic energy, while in the second scenario, the height will be the same as 1.40 m because the total kinetic energy remains constant despite the change in rotational inertia.

PREREQUISITES
  • Understanding of kinetic energy equations: Ek(lin) = 1/2mv² and Ek(rot) = 1/2Iw²
  • Knowledge of gravitational potential energy: Ep = mgh
  • Familiarity with the concept of rotational inertia: I = Σmr²
  • Basic principles of energy conservation in mechanical systems
NEXT STEPS
  • Explore the relationship between linear and rotational kinetic energy in different shapes of objects
  • Investigate the effects of rotational inertia on the motion of rigid bodies
  • Learn about energy conservation principles in vertical motion
  • Study the dynamics of rolling motion and its impact on energy distribution
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of energy conservation and motion dynamics in mechanical systems.

SunshineCat
Messages
4
Reaction score
0

Homework Statement


A hollow ball has a mass of 0.310 kg and radius 0.0340 m. The ball is thrown vertically upwards from rest. It rises through a height of 1.40 m then drops down again. When it is released, it is moving upwards at 5.24 m s−1 and rotating at 2.70 revolutions per second.

For the following two situations, explain whether the height to which the ball rises will be less than, greater than, or the same as 1.40 m.
1) The ball is not rotating, but is given the same linear speed when it is released
2) The ball is solid instead of hollow, but has the same mass and radius. The same amount of total work is done to give the ball its linear and rotational motion, and it has the same angular speed.

Homework Equations


Ek(lin) = 1/2mv^2
Ek(rot) = 1/2Iw^2
Ep = mgh
I = \summr^2

The Attempt at a Solution


1) Since ball is not rotating, Ek(rot) = 0. Ek(lin)=1/2mv^2, so if speed (v) same, Ek(lin) same. Total Ek has decreased.
2) Since ball is solid (mass distributed closer to center) and has same mass, rotational inertia (I) is smaller (I = \summr^2). Since Ek(rot) = 1/2Iw^2, less Ek(rot) is gained. Since work done is the same, same total Ek gained. More of this Ek will be as Ek(lin) than Ek(rot) compared to before.

Where I am stuck is about energy conversion between Ep and Ek. At first I thought that since Ep gained = Ek lost, it won't matter what form that Ek is in (so for Q1, since total Ek is less Ep gained will be less and height is less, and for Q2 since total Ek is same, Ep gained will be same and height is the same).
But when you think about it intuitively, it seems like only Ek(lin) contributes to the gravitational Ep gained, since that's the actual movement of the ball upwards? The Ek(rot) is just the ball spinning.
The only solution I can think of to that is that the Ek(rot) is converted into some other form of Ep which isn't gravitational?
If this is the case (only Ek(lin) contributes to Ep(grav)), then for Q1 it will reach the same height and for Q2 it will reach a higher height?
 
Physics news on Phys.org
Yes, "Ep gained= Ek lost" and the other way around. As the ball rises it will gain potential energy so will lose kinetic energy. However, there is no force that will cause the ball to rotate faster or slower. THAT energy will stay the same so does not even have to be considered.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K