The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  

So, moment of rotational inertia (I) of a cylinder about it axis =

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) =

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) =

k = 0.7560

k ≅ 0.76 m

k = 39.2 N / m

Explanation:

The 200 g block is accelerated by the force of friction between the blocks. Let's use Newton's second law

    N- W = 0

    N = W

    fr = ma

   μ N = ma

   μ mg = ma

   a=μ g

Let's look for the acceleration of the largest block that has oscillatory movement

    x = A cos (w t)

   A = 0.05 m

The maximum acceleration is  cos wt = ±1

    a = A w2

    a = A k / m

We substitute and calculate

  μ g = A k / M

   k = μ g M / A

The mass that performs the oscillation is the mass of the two bodies

   M = m1 + m2

   k = 0.2 9.8 (0.800+ 0.200) /0.05

   k = 39.2 N / m