Question
Question: The radius of gyration of a body about an axis at a distance of 6cm and the centre of mass is 10cm. ...
The radius of gyration of a body about an axis at a distance of 6cm and the centre of mass is 10cm. What is the radius of gyration about a parallel axis through its centre of mass?
Solution
This particular numerical problem can be easily solved by using parallel axis theorem and then substituting the relation between moment of inertia and mass of the body.
Formula used: The parallel axis theorem states that about an axis parallel to the body its moment of inertia through its centre is equal to the sum of the moment of inertia passing through its centre and product of the square of the distance between the two axes times the mass of the body. Mathematically it can be expressed as:
Ir=Ic+mr2
Complete step by step answer:
We know from the parallel axis theorem that,
Ir=Ic+mr2
Now we know that the moment of inertia of a body is equal to the product of the mass of the body and the square of the perpendicular distance to the rotating axis. Thus we can write that:
Ir=m×Kr2
Similarly we can write the same for Ic. That is:
Ic=m×Kc2
Now, substituting these values in the above equation we get:
m×Kr2=m×Kc2+m×r2
Now according to the data given in the numerical problem,
Kr=10cm r=6cm
Now we have to make that radius of gyration (Kc)as the subject of the formula. We can also see that we can divide both sides off the equation by m as it would not change the equality. Thus,
Kc=Kr2−r2
Now substituting the values given in the numerical problem we get:
Kc=102−62 ⇒Kc=64
Calculating this further, we get the radius of gyration that is:
Kc=8cm
Thus, the radius of gyration about a parallel axis passing through its centre of mass is equal to 8cm.
Note: It is always better to use the parallel axis theorem straight away than to use any other alternate method as this simplifies the problem and makes the calculations easier. Usually the radius of gyration with respect to centre of mass is used to find the radius of gyration about any other axis, this problem is based on reverse engineering of the same.