How do I correctly calculate linear speed in a tilted pinball machine problem?

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In summary: The first equation is PEsp = KE, 16 = (0.5)(0.1)(v^2), and got 17.89, for my linear speed, and got that wrong. So to get the correct linear speed, you need to add the kinetic energy of rotation to the final kinetic energy and use the equation \omega = {v \over r} In summary, the discussion is about finding the linear and angular speeds of a pinball as it is fired onto a tilted pinball machine surface. The conversation includes calculations using the formula PEsp = KE, where the kinetic energy of rotation was not initially included but must be added to the final kinetic energy. The correct linear speed can be found by using
  • #1
FlipStyle1308
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A pinball (solid sphere of mass = 0.1 kg, outer radius = 0.1 m) is cocked back 0.8 m on a spring (k = 50 N/m), and fired onto the pinball machine surface, which is tilted. When the pinball is 2.5 m above its initial location (on the way up), find its linear and angular speeds.

I used PEsp = KE, which I got 16 = (0.5)(0.1)(v^2), and got 17.89, for my linear speed, and got that wrong. I know that once I get my linear speed, I plug that into an equation to get angular speed, but I would like to know how to get this linear speed correct. Thanks!
 
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  • #2
FlipStyle1308 said:
I used PEsp = KE, which I got 16 = (0.5)(0.1)(v^2), and got 17.89, for my linear speed, and got that wrong. I know that once I get my linear speed, I plug that into an equation to get angular speed, but I would like to know how to get this linear speed correct. Thanks!
2.5 meters?? Wow, that's a huge pinball machine!:biggrin:

You have forgotten to include the kinetic energy of rotation. You must add to the final kinetic energy [itex] {1 \over 2} I \omega^2[/itex]and use [itex]\omega = {v \over r} [/itex]
 
  • #3
So PEsp = KE, 16 = (1/2)(0.1)(v^2) + (1/2)(0.1)(0.1^2)(v/0.1)?
 
  • #4
FlipStyle1308 said:
So PEsp = KE, 16 = (1/2)(0.1)(v^2) + (1/2)(0.1)(0.1^2)(v/0.1)?
I am confused by your second term! First, you need to compute the moment of inertia I of a sphere (you seemed to have use the mass instead of I!) Also, the (v/r) must be squared. And I don't know what you (0.1)^2 means.
 
  • #5
Okay, my mistake! So is velocity = 12.649 m/s, and angular speed is 126.49 rad/s? I got this wrong...I don't know why.
 
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  • #6
Bump! Anyone able to help me finish off this problem?
 
  • #7
All your numerical values, except the mass seems rather suspicious. A ball with a diameter of 20 centimeters going 2.5 meters up in a pinball machine?
[tex]P_{ela}=P_{grav}+K_{lin}+K_{rot}[/tex]
the elastic potential energy is converted to gravitational potential energy and linear and rotational kinetic energy giving
[tex]kx^2=2mgh+mv^2+I{\omega}^2[/tex]
then use
[tex]v=\omega r[/tex]
to subs for [itex]v[/itex] and solve for [itex]\omega[/itex]
 
  • #8
What is x?
 
  • #9
The first equation in my post states that the stored elastic energy in the spring is converted into the gravitational potential energy of the ball, the linear kinetic energy of the ball and the rotational kinetic energy of the ball.
 

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A pinball machine problem refers to any issue or malfunction that occurs with a pinball machine, such as a ball getting stuck, flippers not working properly, or lights not functioning. It can also refer to more complex technical problems that require repairs.

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