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frank1
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Hello guys, how are you?
An uniform solid sphere, that has 20mm of diameter and 8.4g of mass, its release at a inclined plane with 22cm of lenght. The inclined plane 'ends' in a table in which the ball starts to move horizontally. After leaving the table, that has 92cm of height, the ball touchs the floor in a distance of 0.5m from the extremity of the table. What is the module of the angle (alpha) between the inclined plane and the horizontal (in degrees)? Disregard the friction losses, and consider [tex]g=9.8m/s^2[/tex]
[tex]U=mgh[/tex]
[tex]K=((Mv^2)/2))+(Iw^2)/2[/tex]
[tex]y=yo+voyt+(at^2)/2[/tex]
[tex]x=x0+v0xt+(at^2)/2[/tex]
[tex]I(sphere)=(2/5)MR^2[/tex]
Well, first, here is how I see this question:
http://s9.postimage.org/68azq8mzz/physicsforum.jpg
Well, if I had the angle, I think I'd not be confused, but since what he wants is the angle, it makes the algebra, for me, quite tricky...
At the top of the plane it has an energy of U=m*g*h=(8.4*(10^-3))*9.8*(22*(10^-3)*sinAlpha*) and we can equal this to the K=((Mv^2)/2))+(Iw^2)/2.
Here is my first doubt, i need to isolate the v in this equation?
Because to find the angle, I would use the expression 0.5=voCosAlpha*time
But in order to find the 'time' I need to use the y=yo+voyt+(at^2)/2, but I won't have the angle...
Well, the brief of my doubt is: Am I doing the right 'thinking' here?
Thanks in advance, and a note: I tried to use lattex all over the post, but everytime I use it, it creates huge spaces what makes things quite hard... but someday I learn how to use it properly...
Homework Statement
An uniform solid sphere, that has 20mm of diameter and 8.4g of mass, its release at a inclined plane with 22cm of lenght. The inclined plane 'ends' in a table in which the ball starts to move horizontally. After leaving the table, that has 92cm of height, the ball touchs the floor in a distance of 0.5m from the extremity of the table. What is the module of the angle (alpha) between the inclined plane and the horizontal (in degrees)? Disregard the friction losses, and consider [tex]g=9.8m/s^2[/tex]
Homework Equations
[tex]U=mgh[/tex]
[tex]K=((Mv^2)/2))+(Iw^2)/2[/tex]
[tex]y=yo+voyt+(at^2)/2[/tex]
[tex]x=x0+v0xt+(at^2)/2[/tex]
[tex]I(sphere)=(2/5)MR^2[/tex]
The Attempt at a Solution
Well, first, here is how I see this question:
http://s9.postimage.org/68azq8mzz/physicsforum.jpg
Well, if I had the angle, I think I'd not be confused, but since what he wants is the angle, it makes the algebra, for me, quite tricky...
At the top of the plane it has an energy of U=m*g*h=(8.4*(10^-3))*9.8*(22*(10^-3)*sinAlpha*) and we can equal this to the K=((Mv^2)/2))+(Iw^2)/2.
Here is my first doubt, i need to isolate the v in this equation?
Because to find the angle, I would use the expression 0.5=voCosAlpha*time
But in order to find the 'time' I need to use the y=yo+voyt+(at^2)/2, but I won't have the angle...
Well, the brief of my doubt is: Am I doing the right 'thinking' here?
Thanks in advance, and a note: I tried to use lattex all over the post, but everytime I use it, it creates huge spaces what makes things quite hard... but someday I learn how to use it properly...
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