Question

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4 rpm. What is its kinetic energy?

Answer 1

Answer:

0.092 J

Explanation:

Rotational kinetic energy is:

RE = ½ Iω²

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

For a solid cylinder, I = ½ mr².

RE = ½ (½ mr²) ω²

RE = ¼ mr²ω²

Convert ω from rpm to rad/s.

33.4 rev/min × (2π rad/rev) × (1 min / 60 s) = 3.50 rad/s

Plug in values and solve:

RE = ¼ (3.0 kg) (0.10 m)² (3.50 rad/s)²

RE = 0.092 J

Answer 2

The kinetic energy of a uniform solid cylinder is 0.092 J.

What is kinetic energy?

• The energy possessed by the body because of its velocity or motion.
• It is denoted by K
• K = 1/2 mv^2 .
• Where , m = mass of the object , V = velocity

What is the equation for rotational kinetic energy?

• RE = 1/2 Iω^2
• Where I = moment of inertia , ω = angular velocity

Now , converting 33.4 rotation to rad/s.

           ∴ 33.4 rev/min × (2π rad/rev) × (1 min / 60 s)

           = 3.50 rad/s

Now, For a uniform solid cylinder,

                       I = 1/2 mr².

                     ∴RE = 1/2 (1/2 mr²) ω²

                     ∴RE = 1/4 mr²ω²

Putting the values in the above equation we get,

                    RE = 1/4 (3.0 kg) (0.10 m)² (3.50 rad/s)²

                    ∴RE = 0.092 J

So, The kinetic energy of a uniform solid cylinder is 0.092 J.

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