Problem 54 - College Physics

Question

An astronaut notices that a pendulum that took \(2.50 \mathrm{~s}\) for a complete cycle of swing when the rocket was waiting on the launch pad takes \(1.25 \mathrm{~s}\) for the same cycle of swing during liftoff. What is the acceleration of the rocket? (Hint: Inside the rocket, it appears that \(g\) has increased.)

Step-by-Step Solution

Answer

The acceleration of the rocket is approximately 29.4 m/s².

Step 1: Understand the problem

We need to find the acceleration of the rocket when liftoff occurs. The hint suggests that the effective gravitational field inside the rocket has changed, which affects the pendulum's oscillation period.

Step 2: Recall the formula for the period of a pendulum

The period \( T \) of a simple pendulum is given by \( T = 2\pi\sqrt{\frac{L}{g_{eff}}} \), where \( L \) is the length of the pendulum and \( g_{eff} \) is the effective gravitational acceleration.

Step 3: Define periods at different stages

On the launch pad, the pendulum has a period \( T_1 = 2.50 \, \text{s} \). During liftoff, the pendulum's period is \( T_2 = 1.25 \, \text{s} \). We assume \( g_{eff} = g + a \) where \( a \) is the acceleration of the rocket.

Step 4: Express effective gravity on launch pad

On the launch pad, the pendulum’s effective gravity is \( g_{eff1} = g \). Substitute \( T_1 = 2.50 \, \text{s} \) into the period formula: \( 2.50 = 2\pi\sqrt{\frac{L}{g}} \).

Step 5: Express effective gravity during liftoff

During liftoff, effective gravity is \( g_{eff2} = g + a \). Substitute \( T_2 = 1.25 \, \text{s} \) into the period formula: \( 1.25 = 2\pi\sqrt{\frac{L}{g + a}} \).

Step 6: Set up the equation to solve for acceleration

Since the length \( L \) of the pendulum remains the same, divide the second period equation by the first: \[ \frac{1.25}{2.50} = \sqrt{\frac{g}{g + a}} \] Simplify to \[ \frac{1}{2} = \sqrt{\frac{g}{g + a}} \].

Step 7: Solve for the acceleration \( a \)

Square both sides to remove the square root: \( \left(\frac{1}{2}\right)^2 = \frac{g}{g + a} \). That gives \( \frac{1}{4} = \frac{g}{g + a} \). Solve for \( a \): \( g + a = 4g \Rightarrow a = 3g \).

Step 8: Express the final result in terms of \( g \)

Assuming \( g \approx 9.8 \, \text{m/s}^2 \), the acceleration \( a \) of the rocket is \( a = 3 \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{m/s}^2 \).

Key Concepts

Acceleration

Acceleration is a fundamental concept in physics that describes how the velocity of an object changes with time. It is a vector quantity, meaning it has both a magnitude and a direction. In everyday terms, acceleration can be thought of as how quickly something speeds up or slows down.

• Mathematically, acceleration is defined as the change in velocity per unit of time: \( a = \frac{\Delta v}{\Delta t} \)
• The unit of acceleration in the International System of Units (SI) is meters per second squared (m/s²).
• Acceleration can result from a change in speed, a change in direction, or both.

In the context of the pendulum and rocket problem, we're trying to find out how much the acceleration due to the rocket's motion ("\(a\)") adds to the gravitational acceleration ("\(g\)") inside the rocket during liftoff.
By using the pendulum's period during both rest and liftoff, you can calculate this additional acceleration experienced during the launch.

Effective Gravitational Field

The effective gravitational field refers to the perceived gravitational field experienced by an object. While the true gravitational field is constant, the effective gravitational field can change due to added forces, such as acceleration from rocket liftoff.

• The effective gravitational field inside a rocket is the combined effect of Earth's gravity and the rocket's acceleration.
• This field is expressed as \(g_{eff} = g + a\), where \(g\) is the gravitational acceleration, and \(a\) is the rocket's acceleration.
• The change in the effective gravitational field affects how objects inside the rocket, like the pendulum, behave.

In the astronaut's pendulum problem, at rest, the effective gravitational field is simply \(g\), the acceleration due to gravity. During liftoff, however, it increases to \(g + a\), affecting the pendulum's period of oscillation and allowing us to compute the rocket's acceleration.

Pendulum Period

The period of a pendulum is the time it takes to complete one full cycle of its swing. This period is affected by the length of the pendulum and the gravitational field it is in.

• The formula for the period \( T \) of a simple pendulum is given by:\[ T = 2\pi\sqrt{\frac{L}{g_{eff}}} \]
• "\(L\)" is the length of the pendulum, and "\(g_{eff}\)" is the effective gravitational field.
• A shorter period means the pendulum swings back and forth more quickly.

When the pendulum is on the launch pad, its period is 2.50 seconds. During liftoff, the period changes to 1.25 seconds. This tells us that the effective gravitational field has increased, which is due to the rocket's acceleration.

Rocket Liftoff

Rocket liftoff describes the initial phase of a rocket's journey into space when it starts its ascent from the ground. During liftoff, several physical forces interact to propel the rocket upwards.

• The main force comes from the rocket engines, which provides thrust.
• This thrust creates acceleration, adding to Earth's gravitational pull felt inside the rocket.
• The increased effective gravitational field makes objects inside behave as though gravity has increased.

For the astronaut aboard, it might seem that gravity has become stronger, when in fact, the perceived increase is due to the rocket's upward acceleration. Understanding these dynamics helps explain changes in the pendulum's period during the rocket's launch.