An interstellar transport ship is designed as a cylinder with diameter 5 km. It rotates along its central axis at such a rate that people standing on its inner surface experience a fictitious force equal to Earth’s gravity (9.8 m/s2).
a) At what angular speed is the starship rotating? What is the period of rotation?
b) Consider a coordinate system that has its origin at the surface of the cylinder, and that is fixed with respect to the starship. If someone 2 m tall stands upright, what is the inertial force on the person’s head? What is the centrifugal force on their head? Assume the person’s head has a mass of 10 kg. Give both magnitude and direction for both forces.
c) What is the Coriolis force on a 0.5 kg ball thrown at 40 m/s to the right (see diagram)? Give both magnitude and direction. Find the ratio of this force to the force of “gravity.”
d) What is the Coriolis force on the same ball thrown at 40 m/s out of the page (perpendicular to the axis of the cylinder)? Give both magnitude and direction. Find the ratio of this force to the force of “gravity.”

To tackle the problem of the interstellar transport ship, we need to break it down into manageable parts. Let's start with the first question regarding the angular speed and period of rotation.

Calculating Angular Speed and Period of Rotation

The ship is designed as a cylinder with a diameter of 5 km, which gives it a radius (r) of 2.5 km or 2500 meters. The fictitious force experienced by the people inside is equal to Earth's gravity, which is 9.8 m/s². This fictitious force is actually the result of centripetal acceleration, which can be expressed as:

Fictitious Force (a) = ω² * r

Where:

• ω is the angular speed in radians per second.
• r is the radius in meters.

Setting the fictitious force equal to Earth's gravity, we have:

9.8 m/s² = ω² * 2500 m

Now, we can solve for ω:

ω² = 9.8 m/s² / 2500 m

ω² = 0.00392 s⁻²

ω = √0.00392 s⁻² ≈ 0.0625 rad/s

Next, we can find the period of rotation (T), which is the time it takes for one complete rotation. The relationship between angular speed and period is given by:

T = 2π / ω

Substituting our value for ω:

T = 2π / 0.0625 rad/s ≈ 100.53 seconds

Inertial and Centrifugal Forces on a Person

Now, let’s move on to the second part of the question regarding the forces acting on a person standing upright inside the cylinder. The height of the person is 2 m, and we need to determine the forces acting on their head, which we will assume is at a height of 2 m from the base.

The inertial force (F_inertial) acting on the person's head can be calculated using the same fictitious force formula:

F_inertial = m * a

Where:

• m is the mass of the head (10 kg).
• a is the fictitious acceleration (9.8 m/s²).

Thus:

F_inertial = 10 kg * 9.8 m/s² = 98 N

This force acts outward from the center of the cylinder, which means it points away from the axis of rotation.

The centrifugal force (F_centrifugal) is essentially the same as the inertial force in this rotating frame, so:

F_centrifugal = 98 N

Both forces have the same magnitude of 98 N and point in the same direction, outward from the center of the cylinder.

Coriolis Force on a Thrown Ball

Next, we need to analyze the Coriolis force acting on a 0.5 kg ball thrown at 40 m/s to the right. The Coriolis force can be calculated using the formula:

F_Coriolis = 2 * m * v * ω

Where:

• m is the mass of the ball (0.5 kg).
• v is the velocity of the ball (40 m/s).
• ω is the angular speed (0.0625 rad/s).

Substituting the values:

F_Coriolis = 2 * 0.5 kg * 40 m/s * 0.0625 rad/s

F_Coriolis = 2 * 0.5 * 40 * 0.0625 = 2.5 N

The direction of the Coriolis force is perpendicular to the motion of the ball and the axis of rotation. Since the ball is thrown to the right, the Coriolis force will act upward (in the direction of the rotation's axis).

To find the ratio of this force to the force of gravity:

Ratio = F_Coriolis / F_gravity = 2.5 N / 9.8 N ≈ 0.255

Coriolis Force for a Perpendicular Throw

Finally, let’s calculate the Coriolis force when the ball is thrown perpendicular to the axis of the cylinder (out of the page). The formula remains the same:

F_Coriolis = 2 * m * v * ω

Using the same values:

F_Coriolis = 2 * 0.5 kg * 40 m/s * 0.0625 rad/s = 2.5 N

However, the direction of this force will now be to the left (due to the right-hand rule), perpendicular to both the velocity of the ball and the axis of rotation.

Calculating the ratio to the force of gravity again gives us:

Ratio = F_Coriolis / F_gravity = 2.5 N / 9.8 N ≈ 0.255

In summary, we have calculated the angular speed, period of rotation, inertial and centrifugal forces on a person, and the Coriolis forces for different scenarios. Each of these calculations helps us understand the dynamics of rotating systems in a space environment.