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.

a) Estimate the mass of the bat.

m=.552kg

Estimate the length of the bat

estimated length= .75m

(i) in the middle of the bat (Use your estimate.)

(ii) at the end of the bat (Use your estimate.)

(iii) 5 cm from the end of the handle (Use your estimate.)

(b) Assume a realistic bat, which is narrower at the handle than at the opposite end, and consider the three locations from part (a). (i) The estimate from part (a) for the moment of inertia in the middle of the bat is which of the following?

a) too high, b)approximately equal, c) too low

(ii) The estimate from part (a) for the moment of inertia at the end of the bat is which of the following?

a) too high, b)approximately equal, c) too low

(iii) The estimate from part (a) for the moment of inertia 5 cm from the end of the handle is which of the following?

a) too high, b)approximately equal, c) too low

Answer 1

Let handle d = 25 mm and bat 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