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

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To calculate the linear speed of a pinball in a tilted machine, it's essential to consider both linear and rotational kinetic energy along with gravitational potential energy. The equation PEsp = KE must include the rotational kinetic energy term, which requires the moment of inertia of the sphere. The correct approach involves using the relationship between linear speed and angular speed, v = ωr, to solve for both speeds accurately. There was confusion regarding the values used, particularly the mass and dimensions of the pinball. Properly accounting for all forms of energy will yield the correct linear speed and angular speed.
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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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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 {1 \over 2} I \omega^2and use \omega = {v \over r}
 
So PEsp = KE, 16 = (1/2)(0.1)(v^2) + (1/2)(0.1)(0.1^2)(v/0.1)?
 
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.
 
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.
 
Last edited:
Bump! Anyone able to help me finish off this problem?
 
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?
P_{ela}=P_{grav}+K_{lin}+K_{rot}
the elastic potential energy is converted to gravitational potential energy and linear and rotational kinetic energy giving
kx^2=2mgh+mv^2+I{\omega}^2
then use
v=\omega r
to subs for v and solve for \omega
 
What is x?
 
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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