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[H]Physics help

Blogs > Empyrean
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Empyrean
Profile Blog Joined September 2004
16992 Posts
Last Edited: 2008-10-15 03:14:18
October 15 2008 03:14 GMT
#1
Hey, I hate having to do this, mainly because I personally hate being bothered to help others with homework.

But anyway, I've been on fall break until now and classes resume tomorrow, and I have a physics problem set due in 13 hours that I'm having trouble on. Below is the physics problem and what I have so far. I have most of it done, but I can't finish it. It's an introductory physics course (PHY 53), so it shouldn't be too difficult.

82. A baseball bat has a "sweet spot" where a ball can be hit with almost effortless transmission of energy. A careful analysis of baseball dynamics shows that this special spot is located at the point where an applied force would result in pure rotation of the bat about the handle grip. Determine the location of the sweet spot of the bat shown below. The linear mass density of the bat is given by (0.61 + 3.3 x^2) kg/m, where x is in meters measured from the end of the handle. The entire bat is 0.84m long. The desired rotation point should be 5.0 cm from the end where the bat is held.

[image loading]


Ok, so far, I know there are equal and opposite impulses acting on where the ball hits and on the batter's hand. So MΔv = -J - J'.

J' refers to impulse at the pivot point, J refers to impulse at where the ball hits.

Also, IΔω = angular impulse = -Jd. Also, v = d(CM)ω.
So, Md(CM)Δω= -J - J'.

Thus,
J' = -(1 - (M*d(CM)*d)/I)*J.

So we want to make M*d(CM)*d equal to I, the moment of inertia, so the term is always 0 and thus there is no impulse J' applied to the pivot point. So the sweet spot is where the moment of inertia equals the mass times d(CM) times d.

My problem is, they're looking for a numerical solution. I know I have to find where the center of mass is and the moment of inertia. Problem is, I don't know how. I'm thinking moment of inertia is found with the parallel axis theorem, and I think it might be (1/12)Ml^2 + M(r)^2, but I'm not sure what I'm supposed to put for r. Also I'm a little confused on how to find M from the linear mass density of the bat (because I'm stupid and forgot how to do this from high school).

Finally, I'm not sure how to calculate the center of mass. I think you just integrate along the bat, but again, I'm stupid and forgot how to do this from high school.

So yeah, rare case in which I have the theoretical solution but not the numerical solution.

Any help is appreciated...though after thirteen hours, I'll probably delete this blog post since by then my problem set'll've been due and I probably'll've missed some points on this problem.

*
Moderator
Grobyc
Profile Blog Joined June 2008
Canada18410 Posts
October 15 2008 03:18 GMT
#2
micronesia!!!!!
time to shine!!!!!
If you watch Godzilla backwards it's about a benevolent lizard who helps rebuild a city and then moonwalks into the ocean.
micronesia
Profile Blog Joined July 2006
United States24698 Posts
October 15 2008 09:24 GMT
#3
Uh thanks Grobyc but I don't have time to help him right now... have work :<
ModeratorThere are animal crackers for people and there are people crackers for animals.
Empyrean
Profile Blog Joined September 2004
16992 Posts
October 15 2008 12:27 GMT
#4
Eh, don't worry about it.
Moderator
Mooga
Profile Blog Joined June 2007
United States575 Posts
Last Edited: 2008-10-15 14:51:52
October 15 2008 14:47 GMT
#5
Sorry, but I don't really have time to read your post right now, but here's how you do it:

Draw your FBD. Sum forces in the y-direction and you get O(y) + F = m*a. (I'm calling the force: F)
Where O(y) is at the center of rotation and a= Dcm*(alpha). (alpha is angular acceleration). (Dcm is distance from center of rotation to COM) Since we know O(y)=0 (because we are finding the sweet spot), then this becomes:

F=mcm*alpha ----------------------------------------------- (1)

Now, sum the moments about point O (center of rotation) and you get:

F*d = I*alpha ---------------------------------------------------- (2)

substitute F into eq. 2 and you get : mcm*alpha*d = I*alpha

alpha cancels out to give you d= I / (mcm) --------------------------------------- (3)

since I = m*k^2 the equation 3 becomes:

d= k^2 / Dcm

this will give you the distance that you are trying to find which is the sweet spot. You should know what Dcm is based on the IC they gave you. k is the radius of gyration about the point O (which is the center of rotation).

Edit: fuck, the smiley faces are supposed to be * Dcm (times Dcm)
Mooga
Profile Blog Joined June 2007
United States575 Posts
October 15 2008 17:53 GMT
#6
Alright, so I finally read the rest of your post. I didn't realize that you couldn't find the CoM and I.

To find the CoM:

Integrate the equation for the linear mass density of the bat from 0 to 0.84m. Now you have the total mass of the bat. The center of mass is the location where half of the mass resides. So divide the total mass by 2 and set that equal to your integrated equation. This will give you the location of the CoM.


To find the Mass Moment of Inertia:

By definition, I = integral (r^2) dm.
In this case, r is the distance from your center of rotation to your CoM and dm is the equation you're given.


The radius of gyration k is defined as k = square root(Mass Moment of Inertia / mass of the bat)

I hope this clears any issues up.
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