Translational Velocity Problem (Sphere)

AI Thread Summary
The discussion revolves around calculating the translational speed of a bowling ball as it ascends a vertical rise of 0.76 meters, starting with a speed of 3.5 m/s at the bottom. The key equations used are the conservation of energy principles, where kinetic energy (KE) is converted to potential energy (PE) as the ball rises. The participants realize that the initial kinetic energy is not sufficient to overcome the potential energy at the top of the hill, indicating that the ball cannot reach the top with the given speed. An important correction is noted regarding the need to consider the ball's rotational energy, which affects the final velocity calculation. Ultimately, the ball lacks the necessary energy to ascend the hill without additional factors.
DrDonaldDuck
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Homework Statement


"A bowling ball encounters a .760m verticle 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 verticle 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,,,
 
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  • #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
 
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