Question

note: if possible please write out steps and equations

A solid sphere with mass 1.0-kg and a 16-cm-diameter turns about its diameter at 77 rev/min.

(a) What is its kinetic energy?
J

(b) If an additional 5.0 mJ of energy are supplied to the rotational energy, what is the new angular speed of the ball?
  rev/min

Answer 1

Given

solid sphere with mass M = 1.0 kg ,

diameter d = 16 cm , radius R = 8 cm = 0.08 m
Rotating about the diameter, with angular velocity W = 77 rev/min = 77*2pi/60 rad/s = 8.0634 rad/s

the kinetic energy of the sphere is Rotational kinetic energy

(a)
we know that the Rotational kinetic energy k.e = 0.5*I*w^2
I is moment of inertia of the solid sphere about the diameter , = (2/5)*M*R^2

so the kinetic energy is

   k.e = 0.5*(2/5)*M*R^2*w^2

   k.e = 0.5(2/5)(1.0)(0.08^2)(8.0634^2) J

   k.e = 0.0832 J

---------

(b) if 5.0 mJ of energy supplied to the sphere then the total energy is

   E = 0.0832+0.005 J = 0.0882 J

now from kinetic energy equation

   k.e = 0.5*(2/5)*M*R^2*w^2

   0.0882 = 0.5(2/5)(1.0)(0.08^2)(w^2) J
solving for w

   w = 8.301 rad/s

The new angular speed of the sphere is w = 8.301 rad/s