Question

In a new model of a machine, a spinning solid spherical part of radius R must be replaced by a ring of the same mass which is to have the same kinetic energy. Both parts need to spin at the same rate, the sphere about an axis through its center and the ring about an axis perpendicular to its plane at its center. 1) What should the radius of the ring be in terms of R?

Answer 1

We would compare the formulas for calculating the rotational kinetic energies of a solid sphere and a ring of same mass. The formula for calculating the rotational kinetic energy, RKE of a solid sphere is expressed as

RKE = 1/2Iw^2

where

I is the moment of inertia

w is the angular velocity

The formula for calculating the moment of inertia, I of the solid sphere is

I = 2/5mR^2

where

m = mass of the sphere

R = radius of the sphere

By substituting I = 2/5mR^2 into RKE = 1/2Iw^2, we vabe

RKE = 1/2 *2/5mR^2* w^2 = 1/5mw^2R^2

The formula for calculating the rotational kinetic energy of a ring which rotates about an axis passing through its center and perpendicular to the plane is expressed as

RKE = 1/2mw^2r^2

where

r^2 is the radius

m is the mass

Since the rotational kinetic energy of the sphere and disk would be the same, we would equate both formulas. We have

1/5mw^2R^2 = 1/2mw^2r^2

Since the mass and speed would be the same, they would cancel out on both sides of the equation. Cancelling mw^2 on both sides, we have

1/5R^2 = 1/2r^2

Dividing both sides of the equation by 1/2, we have

r^2 = 1/5R^2/(1/2) = 1/5R^2 * 2

r^2 = 2/5R^2

r^2 = 0.4R^2

Taking the square root of both sides of the equation,

`\begin{gathered} r\text{ = }\sqrt[]{0.4R^(^2)} \\ r\text{ = 0.63R} \end{gathered}`

The radius of the ring in terms of R is 0.63R