Question

A solid sphere, with a mass of 3 kg, is rolling across a horizontal floor without slipping. The sphere has 8.4 J of total kinetic energy and has a center of mass velocity of 2 m/s. Calculate the rotational kinetic energy of the solid sphere.

Answer 1

Answer:

2.4 J

Explanation:

The total kinetic energy of the solid sphere = rotational kinetic energy + translational kinetic energy

KE(total) = KE(rot) + KE(trans)

KE(rot) = KE(total) - KE(trans)

Since its total kinetic energy KE' = 8.4 J and its mass, m = 3 kg and the velocity of its center of mass, v = 2m/s, KE(trans) = 1/2mv² = 1/2 × 3 kg × (2 m/s)² = 3 kg × 2 m²/s² = 6 kgm²/s² = 6 J

So, KE(rot) = KE(total) - KE(trans)

Substituting the values of the variables, we have

KE(rot) = 8.4 J - 6 J

KE(rot) = 2.4 J

So, the rotational kinetic energy of the solid sphere is 2.4 J

Answer 2

The rotational kinetic energy of the solid sphere will be "2.4 J".

Kinetic energy

According to the question,

Total Kinetic energy, KE' = 8.4 J

Mass, m  = 3 kg

Center of mass's velocity, v = 2 m/s

Now,

KE(trans) = $\frac{1}{2}$mv²

By substituting the values,

               = $\frac{1}{2}$  × 3 × (2)²

               = $\frac{1}{2}$ × 3 × 4

               = 6 kgm³/s² or,

               = 6 J  

We know the relation,

→ KE(rot) = KE(total) - KE(trans)

By putting the values,

              = 8.4 - 6

              = 2.4 J

Thus the approach above is correct.  

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