Question

A solid sphere with a moment of inertia of 2.60 kgm�� is rolling along the ground (without slipping) with an angular velocity of 11.1 rad/s. Find (a) its rotational kinetic energy and (b) its translational kinetic energy.

Answer 1

The rotational kinetic energy of the solid sphere is 160.99 J and the translational kinetic energy is 38.12 J.

Step-by-step explanation:

To find the rotational kinetic energy of the solid sphere, we can use the formula:

Krot = (1/2) I ω2

where I is the moment of inertia and ω is the angular velocity.

Plugging in the given values, we get:

Krot = (1/2) (2.60 kgm²) (11.1 rad/s)2 = 160.99 J

To find the translational kinetic energy, we can use the fact that the rolling motion of the sphere consists of both rotational and translational motion. The total kinetic energy is the sum of the rotational and translational kinetic energies:

Ktotal = Krot + Ktrans

Since the sphere is rolling without slipping, the translational velocity can be related to the angular velocity as:

v = ωr

where v is the translational velocity and r is the radius of the sphere. Solving for v, we get:

v = (11.1 rad/s) (r)

Plugging in the given radius, we get:

v = (11.1 rad/s) (0.5 m) = 5.55 m/s

Using the formula for translational kinetic energy:

Ktrans = (1/2) mv2

Plugging in the given mass and translational velocity, we get:

Ktrans = (1/2) (2.60 kg) (5.55 m/s)2 = 38.12 J

Therefore, the translational kinetic energy of the solid sphere is 38.12 J.