How Do You Calculate the Moment of Inertia for a Tennis Racket?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a tennis racket, which consists of a handle and a flat circular disc. The original poster presents a specific configuration with given masses and dimensions, seeking to determine the moment of inertia about an axis through the center of mass.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the moment of inertia equations for a thin rod and a circular disc, considering the parallel axis theorem for the handle. There are questions about the contributions from each part of the racket and how to find the common center of mass.

Discussion Status

Some participants have offered guidance on using the parallel axis theorem and finding the center of mass, while others express confusion about the calculations and seek clarification on techniques applicable to similar problems. Multiple interpretations of the setup are being explored.

Contextual Notes

There is uncertainty regarding the shape of the handle and how it affects the calculations. The original poster's approach to combining the moment of inertia for the handle and disc is under scrutiny, with participants questioning the assumptions made in the setup.

metalInferno
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Homework Statement


a table teennis racket has a handle of mass m and length l and a flat circular disc of radius r and mass m attached to the handle . the moment of inertia of the racket about an axis perpendicular to its plane and passing through its center of mass of is ...?

Homework Equations


i think probably the eqs of MI of a thin rod of length l , and of a circular disc.

The Attempt at a Solution


i thought that by taking out the moment of inertia of rod and circular disc separately i will b able to solve it by later adding the answers . but i ended up with this instead [tex]\frac{ML^2}{3}[/tex] + [tex]\frac{3MR^2}{2}[/tex]

whereas the answer is :
[tex]\frac{ML^2}{12}[/tex] +[tex]\frac{MR^2}{2}[/tex] + [tex]\frac{M(L+2R)^2}{8}[/tex]
 
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Use parallel axis theorem. Isn't anything mentioned about the shape of the handle? Then you have to take it as a thin rod.
 
I don't know how you got your first answer but in the second one the first and second terms are the contributions from the rod and the disk and the last comes from the parallel axis theorem. Find the distance of the center of mass of each from the common center of mass. Hi, Shooting star. Nice medal!
 
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did everything yet the answer does not come!
 
yeah but can u please enlighten me on how 2 to get the common centre of mass . and please state ur technique which i can use for other ques like this as well . thanks for the same . :)
 
metalInferno said:
yeah but can u please enlighten me on how 2 to get the common centre of mass . and please state ur technique which i can use for other ques like this as well . thanks for the same . :)

Pick a coordinate system. Find COM (center of mass) of each. Then the common COM is (COM1*M1+COM2*M2)/(M1+M2).
 
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Suppose the CM lies on the handle at dist x from where the handle meets the disk. Then,

m(r+x) = (m/L)(L-x)[(L-x)/2].

(EDIT.)
 
Last edited:
Dick said:
Hi, Shooting star. Nice medal!

Thank you for noticing.
 

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