Translational Velocity Problem (Sphere)

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Homework Help Overview

The problem involves a bowling ball ascending a vertical rise of 0.760 meters, with an initial translational speed of 3.5 m/s at the bottom. The discussion centers on the relationship between kinetic energy and potential energy as the ball moves up the hill, while ignoring frictional losses.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of kinetic and potential energy equations, questioning the implications of energy conservation as the ball ascends. Some express confusion regarding the role of the factor of 0.5 in the kinetic energy formula and its relation to potential energy. Others attempt to clarify the energy transformations involved.

Discussion Status

The discussion has highlighted various interpretations of energy conservation principles, with participants questioning assumptions about initial and final energies. Some have provided guidance on how to approach the problem, while others have noted potential errors in reasoning related to energy calculations.

Contextual Notes

Participants note that the initial potential energy is considered to be zero, and there is an ongoing debate about the necessity of accounting for rotational energy in the calculations. The discussion reflects a lack of consensus on the correct approach to solving the problem.

DrDonaldDuck
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Homework Statement


"A bowling ball encounters a .760m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass of ball is distributed uniformly. Translational speed of ball =3.5 m/s at the bottom of the rise. Find translational speed at the top."

-Ball is going up a vertical hill (IMPOSSIBRU?)
acceleration is -9.8m/s^2
translational speed = 3.5 m/s at BOTTOM of vertical hill (initial translational speed)
hill is .76m high

Homework Equations


KE=PE
mgh=.5mv^2

The Attempt at a Solution


-m's cancel in equations-
.5(9.8m/s^2)(.760m)=.5(v)^2
v^2 = 14.xxx
v>3.5(3.5 is initial velocity), this is impossible because the translation velocity is supposed to decrease when ball rolls up hill, not increase.
 
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DrDonaldDuck said:

Homework Statement


"A bowling ball encounters a .760m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass of ball is distributed uniformly. Translational speed of ball =3.5 m/s at the bottom of the rise. Find translational speed at the top."

-Ball is going up a vertical hill (IMPOSSIBRU?)
acceleration is -9.8m/s^2
translational speed = 3.5 m/s at BOTTOM of vertical hill (initial translational speed)
hill is .76m high

Homework Equations


KE=PE
mgh=.5mv^2


The Attempt at a Solution


-m's cancel in equations-
.5(9.8m/s^2)(.760m)=.5(v)^2 ##\leftarrow##
v^2 = 14.xxx
v>3.5(3.5 is initial velocity), this is impossible because the translation velocity is supposed to decrease when ball rolls up hill, not increase.

What does the initial ".5" represent in the expression indicated above?

The vertical rise only provides a decrease in kinetic energy as KE is traded for increased PE.
Find the initial KE and the final KE, and thus final velocity.
 
The .5 is in the formula for Kinetic Energy (KE), which is .5mv^2.
 
DrDonaldDuck said:
The .5 is in the formula for Kinetic Energy (KE), which is .5mv^2.
Why is it multiplying a potential energy term on the left hand side?
 
Because when the ball reaches the top, it will have gravitational potential energy.
 
DrDonaldDuck said:
Because when the ball reaches the top, it will have gravitational potential energy.

Sure, but what is that 0.5 doing there? It's not the mass of the ball, it wasn't given. It's not the change in height, because that's 0.760m and is already there to the right of the gravitational acceleration (g = 9.8 m/s2). The kinetic energy change is on the right side of the equals sign, and it has the expected 0.5.
 
gneill said:
Sure, but what is that 0.5 doing there? It's not the mass of the ball, it wasn't given. It's not the change in height, because that's 0.760m and is already there to the right of the gravitational acceleration (g = 9.8 m/s2). The kinetic energy change is on the right side of the equals sign, and it has the expected 0.5.

The 0.5 is part of the universal equation to find Kinetic energy.

Google "Kinetic energy" > Wikipedia entry > "Kinetic energy of rigid bodies" and you will see the equation KE= 0.5mv^2.
 
DrDonaldDuck said:
The 0.5 is part of the universal equation to find Kinetic energy.

Google "Kinetic energy" > Wikipedia entry > "Kinetic energy of rigid bodies" and you will see the equation KE= 0.5mv^2.

The kinetic energy is on the right hand side of your equation. The left hand side is the potential energy (it involves the mass, the acceleration due to gravity, and the change in height).

Your procedure should be to find the change in potential energy due to the change in elevation, and then apply that change in energy to the initial kinetic energy to find the final velocity.
 
My bad. Thanks!
Hold on... I used
KE(initial) + PE(initial) = KE(final) + PE (final)
The initial PE is 0, because the initial height/elevation is 0.
so:
.5m(3.5m/s)^2 = m(9.8)(.760) + .5m(v)^2.
The m's (masses) cancel, but I got
6.125 = 7.448 + .5(v)^2.
This must be wrong as I would get a negative velocity. I tried making "g" into (-9.8m/s^2) but still no luck. This is weird,,,
 
Last edited:
  • #10
Maybe the final velocity is negative?
But then there can't be a square root of a negative number. Help!
 
  • #11
DrDonaldDuck said:
Maybe the final velocity is negative?
But then there can't be a square root of a negative number. Help!

Yup, it looks as though the ball hasn't sufficient energy to make it to the top of the rise.
 
  • #12
Wait.. I found out my error! I also had to take in the rotational energy of the ball into account :D

/marked as solved
 
Last edited:

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