On Earth a baseball player can hit a ball 120 m by giving it an initial angle of 45° to the horizontal. Take the acceleration due to gravity as g=10 ms^-2 Suppose the batter repeats this exercise in a space 'habitat' that has the form of a circular cylinder of radius R = 10 km and has an angular velocity about the axis of the cylinder sufficient to give an apparent gravity of g at radius R. The batter stands on the inner surface of the habitat (at radius R) and hits the ball in the same way as on Earth (i.e., at 45° to the surface), in a plane perpendicular to the axis of the cylinder (see Figure P. 1.45). What is the furthest distance the batter can hit the ball, as measured along the surface of the habitat?

Question

On Earth a baseball player can hit a ball 120 m by giving it an initial angle of 45° to the horizontal. Take the acceleration due to gravity as g=10 ms^-2 Suppose the batter repeats this exercise in a space 'habitat' that has the form of a circular cylinder of radius R = 10 km and has an angular velocity about the axis of the cylinder sufficient to give an apparent gravity of g at radius R. The batter stands on the inner surface of the habitat (at radius R) and hits the ball in the same way as on Earth (i.e., at 45° to the surface), in a plane perpendicular to the axis of the cylinder (see Figure P. 1.45). What is the furthest distance the batter can hit the ball, as measured along the surface of the habitat?

Answer 1

To find the furthest distance the batter can hit the ball on the circular surface of the habitat, we need to consider the effects of gravity and the angular velocity of the habitat.

First, let's calculate the apparent gravity at radius R. Apparent gravity is the centrifugal force experienced by an object rotating in a circular motion. It is given by the formula:

g_apparent = ω^2 * R

Where ω is the angular velocity of the habitat and R is the radius. In this case, the apparent gravity is equal to the acceleration due to gravity on Earth (g = 10 m/s²), so we can set up the equation:

10 m/s² = ω^2 * 10,000 m

Solving for ω, we have:

ω^2 = 1,000 (cancelling out the meters)

ω = √(1,000) = 31.62 rad/s (taking the positive value)

Now, let's consider the ball's motion. We'll assume the initial speed and launch angle remain the same as on Earth.

The horizontal distance the ball travels before hitting the ground (along the surface of the habitat) can be determined using the range formula for projectile motion:

R = (v^2 * sin(2θ)) / g

Where:
R is the range or horizontal distance
v is the initial velocity
θ is the launch angle with respect to the horizontal
g is the acceleration due to gravity

Given that the initial velocity remains the same as on Earth, which is v = 120 m, and the launch angle is 45° (π/4 radians), we can plug in these values along with the acceleration due to gravity on this habitat (g_apparent = 10 m/s²) into the range formula:

R = (120^2 * sin(2(π/4))) / 10

Using a calculator, we find:

R ≈ 172.8 m

Therefore, the furthest distance the batter can hit the ball on the surface of the habitat, as measured along the surface, is approximately 172.8 meters.