A Pinball, Spring and Conservation of Energy Problem

AI Thread Summary
The discussion centers on calculating the linear and angular speeds of a pinball after it is fired from a spring. Initially, 12.25 J of energy is stored in the spring, and the pinball reaches a maximum height of 6.24363 m. When the pinball is 2.5 m above its starting point, its potential energy is expressed as mg₀h, contributing to the total mechanical energy. The relationship between potential and kinetic energy allows for the calculation of linear speed using the equation mg₀h + ½mv² = 12.25 J. To find angular speed, the linear speed is divided by the pinball's circumference, facilitating the conversion.
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A pinball (solid sphere of mass=0.2 kg, outer radius=0.3 m) is cocked back 0.7 m on a spring (k= 50 N/m), and fired onto the pinball machine surface, which is tilted.



c) When the pinball is 2.5 m above its initial location (on the way up), find its linear and angular speeds.



I got the answers to parts a through c as the following: a) 12.25 J of energy is initially stored in the spring
b) 6.24363 is the maximum height the pinball rolls to (above its initial location.) However, I'm having trouble tackling part c..not really sure where to begin.


Anything will help. Thank you so much!
 
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I'll have a go.

When the ball is 2.5 metres above its initial position, it has acquired mg_{0}h joules of potential energy.

The mechanical energy of the ball (which was calculated in part A) is equal to the sum of its kinetic energy and potential energy. Thus,

mg_{0}h+\frac{1}{2}mv^{2}=12.25J

The linear speed can now be calculated from this equation.

To convert from linear speed to angular speed, I believe the linear speed is divided by the circumference (I find it can help to think about the units: metres per second divided by metres per revolution yields revolutions per second).

Hope this helps!
 
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