Max Height Calculation for a Spring-Loaded Pinball on a Tilted Surface

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In summary, the conversation discusses using energy conservation to find the maximum height a pinball will roll to when fired from a spring onto a tilted pinball machine surface. The formula used is 0.5m(vi^2) + 0.5k(yi^2) = 0.5m(vf^2) + 0.5k(yf^2) and the height is found by setting the elastic potential energy of the spring equal to the gravitational potential energy of the ball. Assumptions are made for the use of energy conservation, such as no friction and only elastic potential energy, kinetic energy, and potential energy being involved. The angle of the machine tilt may also be important and the assumption is
  • #1
Nm
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Let's say I have a pinball (solid sphere) with a given mass and radius, and it is held back at a certain distance from equilibrium on a spring. The spring constant k is given, and the ball is then fired onto the pinball machine surface, which is tilted. How would I find the maximum height the pinball rolls to (above its initial location.)? I know I should use energy conservation, but is this the correct formula: 0.5m(vi^2) + 0.5k(yi^2) = 0.5m(vf^2) + 0.5k(yf^2)? Would yf be the same as the height at which the pinball machine is tilted or is it the distance from the spring equilibrium the pinball travels to? Thanks.
 
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  • #2
The potential energy stored in the spring as the ball is pushed and held back is then converted into kinetic energy as the ball is moving up the machine, then into gravitational potential energy as it moves up the surface.

The potential energy os a spring is,
PE_s = 1/2 k*x^2
The gravitational potential energy is,
PE_g = mgh

so if you set these equal to each other,
1/2*k*x^2 = mgh
now solve for the height h (or y, whatever you want to call it)
 
  • #3
Thanks for your help. I have another question: When the pinball is at a certain distance above its initial location (on the way up), how would I find its speed? Do I set KE = PE of the spring?
 
  • #4
in this case,
elastic potential energy of the spring = kinetic energy of the ball + its gravitational potential energy.

assumptions we have to made in order to use the law of conservation of energy:
1. the surface of the machine has no friction.
2.only elastic potential energy, kinetic energy and potential energy are involved. no sound energy is produced.

what i know is that the h in the formula mgh refers to the vertical distance or height. so i think it is important to know the angle the machine is tilted.
 
  • #5
Leong said:
in this case,
elastic potential energy of the spring = kinetic energy of the ball + its gravitational potential energy.
assumptions we have to made in order to use the law of conservation of energy:
1. the surface of the machine has no friction.
2.only elastic potential energy, kinetic energy and potential energy are involved. no sound energy is produced.
what i know is that the h in the formula mgh refers to the vertical distance or height. so i think it is important to know the angle the machine is tilted.
Another thing you assume by saying that the,
"elastic potential energy of the spring = kinetic energy of the ball + its gravitational potential energy." is that the ball is not spinning and/or it is a point mass.
I only bring this up because in the original statement of the problem, Nm said that the ball had a given radius (amongst other things). This almost seems to imply that we are suppose to take the rotation of the ball into account. If this is indeed the case, then not only does the ball have gravitational potential energy, and [normal] translational kinetic energy, but also rotational kinetic energy (the "spin" energy if you will).
 

1. What is the "Simple Pinball Spring Problem"?

The "Simple Pinball Spring Problem" is a classic physics problem that involves a spring and a mass attached to it. The mass is released from rest and oscillates back and forth due to the force of the spring.

2. What are the key factors that affect the motion of the mass in the "Simple Pinball Spring Problem"?

The key factors that affect the motion of the mass in the "Simple Pinball Spring Problem" are the mass of the object, the spring constant, and the amplitude of the oscillations. These factors determine the frequency and period of the oscillations.

3. How can the "Simple Pinball Spring Problem" be solved?

The "Simple Pinball Spring Problem" can be solved using the equation for simple harmonic motion, which is T = 2π√(m/k), where T is the period of the oscillations, m is the mass of the object, and k is the spring constant. This equation can be used to calculate the frequency, period, and other parameters of the motion.

4. What are some real-life applications of the "Simple Pinball Spring Problem"?

The "Simple Pinball Spring Problem" has many real-life applications, such as in the design of suspension systems for vehicles, shock absorbers, and springs in mechanical devices. It is also used in the study of sound waves and in the construction of musical instruments.

5. How does changing the parameters of the "Simple Pinball Spring Problem" affect the motion of the mass?

Changing the parameters of the "Simple Pinball Spring Problem" can have a significant impact on the motion of the mass. For example, increasing the mass will result in a longer period of oscillation, while increasing the spring constant will result in a shorter period. Changing the amplitude can also affect the frequency and period of the motion.

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