Answer in Mechanics | Relativity for Ojugbele Daniel #138997
March 12th, 2023
The moment of inertia of a disc of mass “M” and radius “R” about an axis through its center and perpendicular to its plane is given by
“I=frac{1}{2}MR^2” …………..(1)
In terms of radius of gyration, “I=MK^2” …………….(2)
Here, “I” is the moment of inertia and “K” is the radius of gyration.
From (1) and (2),
“MK^2=frac{1}{2}MR^2\nRightarrow K^2=R^2/2\nRightarrow K=R/sqrt 2”
The radius of the disc is “R=25 cm”
Therefore, the radius of gyration is “K=frac{25 cm}{sqrt 2} = 17.7 cm”
Answer: The radius of gyration of the disc which spins on an axis through its centre and perpendicular to its plane is 17.7 cm.
