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Help Finding Rotational Inertia
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You possess a ball of mass 450 g and radius 4 cm, but with an unknown mass distribution, and thus unknown rotational inertia.Being an inquisitive physics student, you decide to roll the ball down a 35degree incline to find the balls rotational inertia. After placing the ball 75cm vertically up the ramp and releasing, you experimentally measure the balls final speed as 3.03 m/s. What is the balls rotational inertia?
I was given some help with the equation:
mgh = 1/2mv^2 +1/2 Iw^2 = 1/2m(3.03)^2 + 1/2Iw^2
Then that was simplified to: 1/2Iw^2 = mgh - 1/2mv^2
That was then set to this equation: I = 2r^2/v^2(mgh-1/2mv^2)
So would my answer be 2(.04)^2/3.03^2(.450x9.8x.75 - 1/2(.450x)(3.03)^2
That would equal... 4.34 x 10^-4
Am I right? XD please help
I was given some help with the equation:
mgh = 1/2mv^2 +1/2 Iw^2 = 1/2m(3.03)^2 + 1/2Iw^2
Then that was simplified to: 1/2Iw^2 = mgh - 1/2mv^2
That was then set to this equation: I = 2r^2/v^2(mgh-1/2mv^2)
So would my answer be 2(.04)^2/3.03^2(.450x9.8x.75 - 1/2(.450x)(3.03)^2
That would equal... 4.34 x 10^-4
Am I right? XD please help
Answers
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Actually it's quite a clever wee experiment, you always want to know how your balls will roll Minty.
I'd use Split Integral Calculus to determine the moment of inertia Sandra, using the quantum explanaion that work in the dynamic field as well. However, I may come back to this because I don't have the book to check the calculus at work
I'd use Split Integral Calculus to determine the moment of inertia Sandra, using the quantum explanaion that work in the dynamic field as well. However, I may come back to this because I don't have the book to check the calculus at work
This is actually just for review for an advanced physics class I'm taking next year, all of this is independent study guys XD. No teacher involved just research I've done through these problems as I work through this workbook . Basically the final equation I come up with is I = 2radius^2/velocity^2 (massgravityheight - .5massvelocity^2)
I figure I just plug in all the information then but I'm unsure and this book doesn't have an answer key in it :/ I can just ask a professor at the university later if I can't work through it.
I figure I just plug in all the information then but I'm unsure and this book doesn't have an answer key in it :/ I can just ask a professor at the university later if I can't work through it.
Okay,
what I was getting at is that
I = 2r^2/(v^2(mgh-1/2mv^2))
gives a different result to
I = (2r^2/v^2) * (mgh-1/2mv^2)
Top one gives 2.8068x10^-4 because 2r^2 is divided by everything else
Bottom one gives 4.328x10^-4 because 2r^2 is multiplied up by the mgh portion
Note:- this is because I can't see how the equation is properly written and can't work out, for myself, what rearrangements went on in the unexplained jump from w^2 to the equation above. (maths 'identities' always tripped me up. I'm guessing w^2 is one of those. If it's the physics then it's a part I've forgotten). :-/
what I was getting at is that
I = 2r^2/(v^2(mgh-1/2mv^2))
gives a different result to
I = (2r^2/v^2) * (mgh-1/2mv^2)
Top one gives 2.8068x10^-4 because 2r^2 is divided by everything else
Bottom one gives 4.328x10^-4 because 2r^2 is multiplied up by the mgh portion
Note:- this is because I can't see how the equation is properly written and can't work out, for myself, what rearrangements went on in the unexplained jump from w^2 to the equation above. (maths 'identities' always tripped me up. I'm guessing w^2 is one of those. If it's the physics then it's a part I've forgotten). :-/
I note how the Yahoo best answer treats the phrase "75cm vertically up the ramp" as meaning
"h = 0.75*sin35° = 0.43 m"
Being pedantic, I would call that 75cm ALONG the ramp which, as calculated, is 43cm above the ground surface. The 75cm affects the velocity calculation and h affects the PE calculation.
When they say "read the question carefully", they really mean it!
"h = 0.75*sin35° = 0.43 m"
Being pedantic, I would call that 75cm ALONG the ramp which, as calculated, is 43cm above the ground surface. The 75cm affects the velocity calculation and h affects the PE calculation.
When they say "read the question carefully", they really mean it!
Hmm, puzzler this one, Like Hypognosis I'm certain there's something wrong or missing with your maths and or your physics here. There is too big a jump from you first equation to your second with out any explanation where w comes from or disappears to.
I assume it's weight (ie mass x gravity) we're talking about, i'm afraid I could come up with a sensible answer using what you given us. Additionally as Hypognosis says "the angled plane" adds significantly to the moment of inertia with a non-constant density or mass distributed ball.... the steeper the plane the greater the influence of gravity.
I had a look from a slightly different perspective: defining the moment of inertia with a non consistent mass or density is beyond me with out digging through some serious physics notes that are not readily to hand. What is to hand was some notes I had on Rotational Moments of inertia in a solid sphere... I'll apologise for the hand written images...ascii keys are murder!!
Anyway looking at the moment of inertia of a solid sphere of constant density about an axis through its centre of mass. You can determine this by summing the individual moments of inertia of all the narrow discs that create the sphere. If the surface of the ball is defined by the equation
http://
(I thinks its part of Fermat's theorem on pythagorean triples)
Anyway the cross section of the sphere through axis Z is represented by
http://
So the moment of inertia of the sphere or ball is the sum of the moments of inertia of the discs along the z-axis and can be solved by the following equation.........
http://
Anyway Apologies if this is a bum steer.... it's been a while since I looked at anything like that.... as it says at the bottom of the notes.... I THINK!!!
Good Luck......
I assume it's weight (ie mass x gravity) we're talking about, i'm afraid I could come up with a sensible answer using what you given us. Additionally as Hypognosis says "the angled plane" adds significantly to the moment of inertia with a non-constant density or mass distributed ball.... the steeper the plane the greater the influence of gravity.
I had a look from a slightly different perspective: defining the moment of inertia with a non consistent mass or density is beyond me with out digging through some serious physics notes that are not readily to hand. What is to hand was some notes I had on Rotational Moments of inertia in a solid sphere... I'll apologise for the hand written images...ascii keys are murder!!
Anyway looking at the moment of inertia of a solid sphere of constant density about an axis through its centre of mass. You can determine this by summing the individual moments of inertia of all the narrow discs that create the sphere. If the surface of the ball is defined by the equation
http://
(I thinks its part of Fermat's theorem on pythagorean triples)
Anyway the cross section of the sphere through axis Z is represented by
http://
So the moment of inertia of the sphere or ball is the sum of the moments of inertia of the discs along the z-axis and can be solved by the following equation.........
http://
Anyway Apologies if this is a bum steer.... it's been a while since I looked at anything like that.... as it says at the bottom of the notes.... I THINK!!!
Good Luck......
I agree with your answer Sandra, well more or less, I get 4.32828 x 10^-4 Kgm^2
The equations are just Conservation of energy equations so that:
Potential Energy =Kinetic Energy of the ball due to linear motion + Kinetic energy of ball due to rotation,
and the question assumes that the ball's rotation is such that the velocity of the ball at its outside edge is the same as its linear velocity.
The equations are just Conservation of energy equations so that:
Potential Energy =Kinetic Energy of the ball due to linear motion + Kinetic energy of ball due to rotation,
and the question assumes that the ball's rotation is such that the velocity of the ball at its outside edge is the same as its linear velocity.
@vascop
you typo'd 45cm. OP says
"After placing the ball 75cm vertically up the ramp… "
@Slush't
You could have saved yourself all the writing and photos just by referring to the equivalent wikipedia page which is practically word-for-word the same. (don't take that the wrong way - one of the page's editors must have the same textbooks or some't like that)
The OP actually says "unknown rotational inertia", not "unknown mass distribution", so she can skip all the complicated maths.
you typo'd 45cm. OP says
"After placing the ball 75cm vertically up the ramp… "
@Slush't
You could have saved yourself all the writing and photos just by referring to the equivalent wikipedia page which is practically word-for-word the same. (don't take that the wrong way - one of the page's editors must have the same textbooks or some't like that)
The OP actually says "unknown rotational inertia", not "unknown mass distribution", so she can skip all the complicated maths.
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