A sphere of uniform density with mass 22 kg and radius 0.7 m is spinning, making one complete revolution every 0.5 s. The center of mass of the sphere has a speed of4 m?s(a) What is the rotational kinetic energy Of the sphere? (b) What is the total kinetic energy of the sphere?

Short Answer

Expert verified

(a) The rotational kinetic energy is, 340.46 J .

(b) The total kinetic energy is, 516.46 J.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The speed of sphere is, V=4 m/s .
  • The mass of sphere is, m=22kg
  • The time of every revolution complete is, T=0.5 s
  • The radius of the sphere is, r=0.7 m
02

Significance of total kinetic energy

The total kinetic energy of an object or system is equal to the sum of the kinetic energy from each type of motion.

03

(a): Determination of the rotational kinetic energy

The relation of rotational kinetic energy is expressed as,

Krot=12Iω2 ...(i)

Here Krotis the rotational kinetic energy, ωis the angular speed and Iis the moment of inertia.

The moment of inertia is expressed as,

I=25m2

And the relation of angular speed is expressed as,

ω=2πT

Here Tis the period of revolution.

Substitute the value of Iand Tin the equation (i).

Krot=1225mr22πT2=15mr22πT2

Substitute 0.5 s for T, 22 kg for m and 0.7 m for r in the above equation

Krot=15×22kg×0.7m22π0.5s2=340.46J

Hence the rotational kinetic energy is, 340.46 J.

04

(e): Determination of the total kinetic energy

The relation of the total kinetic energy is expressed as,

Ktot=Ktrans+Krot ...(ii)

Here Ktransis the translational kinetic energy. It is expressed as,

Ktrans=12mv2

Substitute 4 m/s for v , and 22 kg for min the above equation.

Ktrans=12×22kg×4m/s2=176J

Substitute the value of Ktransand Krotin the equation (ii).

Krot=176J+340.46J=516.46J

Hence the total kinetic energy is, 516.46J.

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