A 725 g metal wire is bent into the shape of a hoop 58.0 cm in diameter. Six wire spokes, each of mass 113 g, are added from the center of the hoop to the rim. What is the moment of inertia of this object about an axis perpendicular to it through its center?
2007-10-23 18:19:45 · 4 answers · asked by KT 1 in Science & Mathematics ➔ Physics
Moment of inertia is linearly additive, so we can calculate the moment of inertia of the hoop and spokes separately.
The moment of inertia for a hoop is mr^2. If the diameter is 58.0, the radius is 29.0 cm, or .29 meters.
The mass is .725 kg, so the moment is (.725)(.29)(.29)= 0.0609725. The problem gives 3 significant figures, so 0.0610.kgm^2.
Think of the six spokes as 3 rods, each with a length of 58.0 cm. The moment of inertia for a rod rotated about its center is 1/3mr^2. In this case that means each rod has a moment of inertia of (1/3)(.113)(.29)^2, so all 3 together have a moment of (.113)(.29)^2 = 0.0095033. There are 3 significant digits given, so 0.00950 is the most precise answer possible.
Adding these two values together gives a total moment of inertia of 0.0705 kgm^2
2007-10-23 18:35:55 · answer #1 · answered by David Zukertort Rudel 3 · 0⤊ 0⤋
Moment of inertia (I) is additive, so find the moments of the individual pieces and add them up. The hoop is easy. It has a radius of (58 cm)/2 = 29 cm, and the moment is:
I_hoop = m*r^2 = (725g) * (29 cm)^2
The spokes are identical. Each is a thin rod of length r. The moment around an endpoint is:
I_spoke = (1/3)*m*r^2 = (1/3) * (113 g) * (29 cm)^2
Now, add up the hoop and six spokes.
2007-10-23 19:02:48 · answer #2 · answered by husoski 7 · 0⤊ 0⤋
You can find a page of formulas for the moments of inertia of various shapes on wikipedia (www.wikipedia.org). Search for "list of moments of inertia".
For your problem, you'll want to add the value for the hoop rotating about its center with six times the value for a rod rotating about its end. The hoop is m*r^2, where m is mass and r is radius, and the rod about its end is 1/3 * m * L^2, where L is its length (solve for L by find the hoop diameter, 58=pi*r)
2007-10-23 18:34:53 · answer #3 · answered by ansrdog 4 · 0⤊ 0⤋
cylinder has m.o.i Mr^2/2 since u have 2 cylinders involved fused side to side u have the libertry to add in that expression r is radius of the cylinder
2016-04-10 01:35:30 · answer #4 · answered by ? 4 · 0⤊ 0⤋
well i have to study that topic again in detail.
2007-10-23 18:30:21 · answer #5 · answered by Anonymous · 0⤊ 0⤋