Monday, 15 September 2014

MOMENT OF INERTIA , ROTATIONAL PROBLEMS AND SOLUTIONS


Problem 1:
What is the moment of inertia of a hoop of mass M and radius R rotated about a cylinder axis, as shown below?
A hoop of radius R
Fortunately, we do not need to use calculus to solve this problem. Notice that all the mass is the same distance R from the axis of rotation. Thus we do not need to integrate over a range, but can calculate the total moment of inertia. Each small element dm has a rotational inertia of R 2 dm , where r is constant. Summing over all elements, we see that I = R 2  dm = R 2 M . The sum of all the small elements of mass is simply the total mass. This value for I of MR 2agrees with experiment, and is the accepted value for a hoop.
Problem 2:
What is the rotational inertia of a solid cylinder with length L and radius R , rotated about its central axis, as shown below?
A cylinder being rotated about its axis
To solve this problem we split the cylinder into small hoops of mass dm , and width dr :
A cylinder being rotated about its axis, shown with a small element of mass from the cylinder
This small element of mass has a volume of (2Πr)(L)(dr) , where dr is the width of the hoop. Thus the mass of this element can be expressed in terms of volume and density:
dm = ρV = ρ(2ΠrLdr)
We also know that the total volume of the entire cylinder is given by: V = AL = ΠR2 L . In addition, our density is given by the total mass of the cylinder divided by the total volume of the cylinder. Thus:
ρ =  = 
Substituting this into our equation for dm ,
dm =  = 2rdr
Now that we have dm in terms of r , we simply have to integrate over all possible values of r to get our rotational inertia:

I= r 2 dm 
 = 2r 3 dr 
 =[r 4/2]0 R 
 = 

Thus the rotational inertia of a cylinder is simply  . Once again, it has the form of kMR 2 , where k is some constant less than one.