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**

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

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