A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (kₛₚₕ: k_(cyl)) is 2: √x. The value of x is ____________ .

Solution

Physics sir · 3 steps

Radius of gyration comes straight from I = Mk², so compare the two moments of inertia about their own axes.

1. Define k. The radius of gyration k is defined by I = Mk², so k = √(I/M). When rolling, each body spins about the axis through its centre.
2. Each body. Solid sphere: I = 2/5MR², so kₛₚₕ = R√(2/5). Solid cylinder about its own axis: I = 1/2MR², so k_(cyl) = R√(1/2).

   kₛₚₕ/k_(cyl) = √((2/5)/(1/2)) = √(4/5) = 2/√5

3. So. The ratio is 2 : √5, so x = 5.

The common slip: 10. Taking the cylinder's moment of inertia as MR² (that is a ring or hollow cylinder) gives a ratio √(2/5) = 2/√10, so x = 10. A solid cylinder has 1/2MR².

The idea to keep. Radius of gyration is just √(I/M). Once you know the standard moments of inertia, any ratio of k values is the square root of the ratio of the fractions in front of MR².

Answer from the NTA final answer key. Solution by PrepWiser.