Question

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890rad/s2. What is the kinetic energy of the shell after it has turned through 6.00 rev?

Answer 1

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

Answer 2

The kinetic energy of the shell after it has turned through 6.00 rev is 8.91J.

How to calculate kinetic energy?

The kinetic energy can be calculated using the following:

• mass, m = 8.20 kg
• radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement can be calculated as follows:

θ = 6 x 2 x π = 37.68 rad

angular acceleration is calculated as follows:

α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Using the third equation of motion as follows:

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy = ½Iw²

K.E = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

Therefore, the kinetic energy of the shell after it has turned through 6.00 rev is 8.91J.

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