- #1
SiliconCPU
- 8
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I have a question regarding the Rolling Motion equation.
[tex](1/2)Mv^2 + (1/2)I\omega^2 + Mgy[/tex]
(Where M=mass, v=velocity, I=moment of inertia,ω=angular velocity, g=force due to gravity and y = vertical distance)
The problem I'm having involves the direction of the acceleration due to gravity. I've seen the equation written as...
[tex](1/2)Mv^2 + (1/2)I\omega^2 = Mgy[/tex]
Are they assuming acceleration due to gravity is downward? Hence, adding Mgy to both sides to produce the above results?
I guess the problem I'm having is distinguishing when to make the acceleration due to gravity positive/negative (in Work/Energy related problems).
[tex](1/2)Mv^2 + (1/2)I\omega^2 + Mgy[/tex]
(Where M=mass, v=velocity, I=moment of inertia,ω=angular velocity, g=force due to gravity and y = vertical distance)
The problem I'm having involves the direction of the acceleration due to gravity. I've seen the equation written as...
[tex](1/2)Mv^2 + (1/2)I\omega^2 = Mgy[/tex]
Are they assuming acceleration due to gravity is downward? Hence, adding Mgy to both sides to produce the above results?
I guess the problem I'm having is distinguishing when to make the acceleration due to gravity positive/negative (in Work/Energy related problems).
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