Calculate the moment of inertia of a baseball bat for the following cases. (a) A
ID: 2300960 • Letter: C
Question
Calculate the moment of inertia of a baseball bat for the following cases.
(a) Assume the bat is a wooden rod of uniform diameter and take the pivot point to be in the middle of the bat, at the end of the bat, and 5 cm from the end of the handle, where a batter who "chokes up" on the bat would place his hands. Hint: Consider the bat as two rods held together at the pivot point. Estimate the mass of the bat.
kg
Estimate the length of the bat?
m
(i) in the middle of the bat (Use your estimate.)?
kg
Explanation / Answer
Let handle dia d = 25 mm and bat dia D = 50 mm
Let handle length l = 250 mm and bat length L = 500 mm.
Density of wood, rho = 500 kg/m^3
a)
Total volume V = volume of handle + volume of bat
= pi/4*d^2*l + pi/4*D^2 *L
= 3.14 / 4 * 0.025^2 * 0.25 + 3.14 /4 * 0.05 ^2 * 0.5
= 0.0011 m^3
Mass of bat = rho*V
= 500*0.0011
= 0.552 kg
Mass of handle m = 500* 3.14 / 4 * 0.025^2 * 0.25 = 0.061 kg
Mass of bat portion M = 0.552 - 0.061 = 0.491 kg
b)
Length = Length of handle + Length of bat = 250 + 500 = 750 mm
c)
i)
Mid point = 750/2 = 375 mm from end of bat.
MOI of rod pivoted about its end = ML^2 /3
MOI of handle about its c.g = ml^2 /12 = 0.061*0.25^2 /12 = 3.177*10^-4 kg-m^2
By parallel axis theorem, MOI of handle about pivot = 3.177*10^-4 + 0.061*(0.375 - 0.25/2)^2 = 0.00413 kg-m^2
MOI of bat portion about its c.g = 0.491*0.5 ^2 /12 = 0.01023 kg-m^2
By parallel axis theorem, MOI of bat portion about pivot = 0.01023 + 0.491*(0.375 - 0.5/2)^2 = 0.0179 kg-m^2
Total MOI about pivot = 0.00413 + 0.0179 = 0.022 kg-m^2
ii)
MOI of handle = 0.061*0.25^2 /3 = 0.00127 kg-m^2
By parallel axis theorem, MOI of bat portion about pivot = 0.01023 + 0.491*(0.75 - 0.5/2)^2 = 0.133 kg-m^2
Total MOI = 0.00127 + 0.133 = 0.13425 kg-m^2
iii)
By parallel axis theorem, MOI of handle about pivot = 3.177*10^-4 + 0.061*(0.05 - 0.25/2)^2 = 0.00066 kg-m^2
By parallel axis theorem, MOI of bat portion about pivot = 0.01023 + 0.491*(0.75 - 0.5/2 - 0.05)^2 = 0.1097 kg-m^2
Total MOI = 0.00066 + 0.1097 = 0.1103 kg-m^2
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