Problem 58 - University Physics with Modern Physics

Question

The position of a \(2.75 \times 10^{5} \mathrm{N}\) training helicopter under test is given by \(\vec{r}=\left(0.020 \mathrm{m} / \mathrm{s}^{3}\right) t^{3} \hat{\imath}+(2.2 \mathrm{m} / \mathrm{s}) f \hat{\jmath}-\left(0.060 \mathrm{m} / \mathrm{s}^{2}\right) t^{2} \hat{k}\) Find the net force on the helicopter at \(t=5.0 \mathrm{s}\)

Step-by-Step Solution

Answer

\( \vec{F} = 16836.73 \hat{\imath} - 3367.35 \hat{k} \text{ N} \) at \( t = 5.0 \text{ s} \).

Step 1: Identify the Position Vector Components

The position vector \( \vec{r} \) is given by three components: \( x = (0.020 \text{ m/s}^3) t^3 \), \( y = (2.2 \text{ m/s}) t \), and \( z = -(0.060 \text{ m/s}^2) t^2 \). These will be differentiated to find the velocity and acceleration.

Step 2: Differentiate Position to Find Velocity

Differentiate each component of the position vector with respect to time \( t \) to find the velocity vector \( \vec{v} \):\( v_x = \frac{d}{dt}[(0.020 \text{ m/s}^3) t^3] = (0.060 \text{ m/s}^3) t^2 \ v_y = \frac{d}{dt}[(2.2 \text{ m/s}) t] = 2.2 \text{ m/s} \ v_z = \frac{d}{dt}[-(0.060 \text{ m/s}^2) t^2] = -0.12 \text{ m/s}^2 t \).

Step 3: Differentiate Velocity to Find Acceleration

Differentiate each component of the velocity vector to find the acceleration vector \( \vec{a} \):\( a_x = \frac{d}{dt}[(0.060 \text{ m/s}^3) t^2] = (0.12 \text{ m/s}^3) t \ a_y = \frac{d}{dt}[2.2 \text{ m/s}] = 0 \text{ m/s}^2 \ a_z = \frac{d}{dt}[-0.12 \text{ m/s}^2 t] = -0.12 \text{ m/s}^2 \).

Step 4: Evaluate Acceleration at t = 5.0 s

Substitute \( t = 5.0 \) seconds into each component of the acceleration vector:\( a_x = (0.12 \text{ m/s}^3)(5.0) = 0.60 \text{ m/s}^2 \ a_y = 0 \text{ m/s}^2 \ a_z = -0.12 \text{ m/s}^2 \). The acceleration vector at \( t = 5.0 \) s is \( \vec{a} = 0.60 \hat{\imath} \text{ m/s}^2 - 0.12 \hat{k} \text{ m/s}^2 \).

Step 5: Calculate Net Force Using Newton's Second Law

Newton's Second Law states \( \vec{F} = m \vec{a} \). First, calculate the helicopter's mass: \( m = \frac{2.75 \times 10^5 \text{ N}}{9.8 \text{ m/s}^2} = 28061.22 \text{ kg} \).The net force is then:\( \vec{F} = 28061.22 \text{ kg} \times (0.60 \hat{\imath} - 0.12 \hat{k}) \text{ m/s}^2 \ = 16836.73 \hat{\imath} - 3367.35 \hat{k} \text{ N} \).

Key Concepts

Position Vector

In physics, a position vector is a fundamental concept that describes the position of a point in space relative to a reference origin. It is usually denoted as \( \vec{r} \) and expressed in terms of its components along the Cartesian coordinates: x, y, and z.
The example given in this exercise specifies the position vector \( \vec{r} = (0.020 \, \text{m/s}^3) \, t^3 \hat{\imath} + (2.2 \, \text{m/s}) \, t \hat{\jmath} - (0.060 \, \text{m/s}^2) \, t^2 \hat{k} \).

• The coefficient terms (like \(0.020 \, \text{m/s}^3\)) scale the dimensions of the vector components.
• The powers of \( t \) indicate how the position changes over time.
• Each unit vector \( \hat{\imath}, \hat{\jmath}, \hat{k} \) represents direction along the x, y, and z axes, respectively.

Understanding how to represent position vectors and interpret their components is crucial for determining motion in physics.

Velocity Vector

The velocity vector provides the rate of change of an object's position with respect to time. It essentially tells us how fast and in which direction an object is moving.
To find the velocity vector, we differentiate the position vector with respect to time. For the exercise:

• \( v_x = (0.060 \, \text{m/s}^3) \, t^2 \) from differentiating \( x \)-component.
• \( v_y = 2.2 \, \text{m/s} \) from differentiating \( y \)-component as it is linear with respect to time.
• \( v_z = -0.12 \, \text{m/s}^2 \, t \) from differentiating \( z \)-component.

The resulting velocity vector \( \vec{v} = (0.060 \, \text{m/s}^3) \, t^2 \hat{\imath} + 2.2 \, \text{m/s} \hat{\jmath} - 0.12 \, \text{m/s}^2 \, t \hat{k} \) illustrates the velocities in each direction, showing how they vary over time.

Acceleration Vector

An acceleration vector quantifies how an object's velocity changes over time. It's the derivative of the velocity vector with respect to time, reflecting both magnitude and direction changes.
To calculate the acceleration for each component:

• \( a_x = (0.12 \, \text{m/s}^3) \, t \) is derived from the \( x \)-component of velocity.
• \( a_y = 0 \, \text{m/s}^2 \) as \( v_y \) is constant and independent of \( t \).
• \( a_z = -0.12 \, \text{m/s}^2 \) comes from the \( z \)-component of velocity.

When evaluated at \( t = 5.0 \, \text{s} \), the acceleration vector becomes \( \vec{a} = 0.60 \, \text{m/s}^2 \hat{\imath} - 0.12 \, \text{m/s}^2 \hat{k} \), displaying specific accelerations in x and z directions.

Mass Calculation

Mass is a measure of the amount of matter in an object and plays a crucial role in mechanics, as it influences how forces affect motion.
The exercise determines mass from the weight equation:\( m = \frac{\text{Weight}}{g} \)
Here,

• The weight of the helicopter is \( 2.75 \times 10^5 \, \text{N} \).
• The acceleration due to gravity \( g \) is \( 9.8 \, \text{m/s}^2 \).

Thus, mass \( m = \frac{2.75 \times 10^5}{9.8} \approx 28061.22 \, \text{kg} \), essential for calculating forces according to Newton's Second Law.

Net Force

Newton's Second Law relates force to mass and acceleration with the equation \( \vec{F} = m \vec{a} \).
In the exercise setup:

• Mass was calculated to be approximately \( 28061.22 \, \text{kg} \).
• Acceleration vector at \( t = 5.0 \, \text{s} \) is \( \vec{a} = 0.60 \, \text{m/s}^2 \hat{\imath} - 0.12 \, \text{m/s}^2 \hat{k} \).

By applying the components to the formula, the net force becomes \( \vec{F} = 28061.22 \, \text{kg} \times (0.60 \hat{\imath} - 0.12 \hat{k}) \, \text{m/s}^2 \).
This results in a net force of \( 16836.73 \hat{\imath} - 3367.35 \hat{k} \,\text{N} \), indicating forces applied in the x and z directions. This concept demonstrates how forces interact with an object's inertia to produce motion.