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Chapter 8: Problem 59

A radio-controlled model airplane has a momentum given by \([(-0.75 kg \cdot m/s^3)t^2\) + \((3.0 kg \cdot m/s)]\hat{\imath}\) + \((0.25 kg \cdot m/s^2)t\hat{\jmath}\). What are the \(x\)-, \(y\)-, and \(z\)-components of the net force on the airplane?

Short Answer

Expert verified
Net force components: \(F_x = -1.50t\), \(F_y = 0.25\), \(F_z = 0\).

Step by step solution

01

Understand Momentum Expression

The given momentum of the airplane is \(p(t) = [(-0.75 kg \cdot m/s^3)t^2 + (3.0 kg \cdot m/s)]\hat{\imath} + (0.25 kg \cdot m/s^2)t\hat{\jmath}\). The momentum is expressed as a vector with components in the \(x\)-direction and \(y\)-direction.
02

Know the Relationship Between Force and Momentum

According to Newton's second law of motion, the force exerted on an object is the derivative of the momentum with respect to time: \(F = \frac{dp}{dt}\).
03

Differentiate Momentum to Find Force in x-direction

To find the component of net force in the \(x\)-direction, differentiate the \(x\)-component of momentum: \[p_x(t) = (-0.75 kg \cdot m/s^3)t^2 + (3.0 kg \cdot m/s)\implies F_x = \frac{d}{dt}[(-0.75)t^2 + 3]\cdot \hat{\imath}= -1.50 kg \cdot m/s^3 \cdot t\].
04

Differentiate Momentum to Find Force in y-direction

To find the component of net force in the \(y\)-direction, differentiate the \(y\)-component of momentum: \[p_y(t) = (0.25 kg \cdot m/s^2)t \implies F_y = \frac{d}{dt}[0.25t]\cdot \hat{\jmath} = 0.25 kg \cdot m/s^2 \].
05

Identify Force in z-direction

There is no \(z\)-component in the given momentum expression, so the force in the \(z\)-direction is \(F_z = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is a fundamental concept in physics that measures the quantity of motion an object possesses. It is a vector, which means it has both a magnitude and a direction. The formula for momentum is given by:\[ p = mv \]where \( p \) represents momentum, \( m \) is the mass of the object, and \( v \) is the velocity. In the context of the exercise, momentum is expressed with components in the \( x \)- and \( y \)-directions, depicted as:
  • \(-0.75 kg \cdot m/s^3 \cdot t^2\) in the \( x \)-direction
  • \(3.0 kg \cdot m/s\) in the \( x \)-direction
  • \(0.25 kg \cdot m/s^2 \cdot t\) in the \( y \)-direction
The \( z \)-direction does not have a component, so it remains zero. Understanding momentum lays the foundation to calculate the forces acting on the object, using differentiation to ascertain changes over time.
Force Calculation
Force calculation, according to Newton's second law of motion, involves determining the derivative of momentum with respect to time. This law is mathematically represented as:\[ F = \frac{dp}{dt} \]This equation indicates that force is essentially the rate of change of an object's momentum. When analyzing the model airplane, we need to differentiate each component of the momentum:
  • In the \( x \)-direction: Differentiating \((-0.75 kg \cdot m/s^3) t^2 + (3.0 kg \cdot m/s)\) yields an \( x \)-component of force \(-1.50 kg \cdot m/s^3 \cdot t\).
  • In the \( y \)-direction: Differentiating \((0.25 kg \cdot m/s^2)t\) gives a constant \( y \)-component of force \(0.25 kg \cdot m/s^2\).
  • In the \( z \)-direction: Since there is no momentum component initially, \( F_z = 0\).
This application of calculus to determine force reinforces the direct connection between momentum and force, where changes in momentum exemplify reflected changes in the applied force.
Differentiation in Physics
Differentiation in physics is a powerful tool used to understand how various quantities change over time. In this context, when we differentiate momentum, as used for the model airplane, we glean insights into the force required to initiate or adjust its motion. The differentiation process allows you to:
  • Measure instantaneous rates of change, crucial for calculating forces from momentum.
  • Apply calculus concepts such as derivatives to real-world physics problems.
  • Determine components of force along different axes, providing a comprehensive view of forces acting on an object.
By applying the derivative, you obtain the instantaneous force applied to an object at a specific moment in time. This analytical technique showcases how momentum translates into force, offering clearer insights into motion dynamics and facilitating more accurate predictions about an object's behavior under varying conditions.

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Most popular questions from this chapter

A small rocket burns 0.0500 kg of fuel per second, ejecting it as a gas with a velocity relative to the rocket of magnitude 1600 m/s. (a) What is the thrust of the rocket? (b) Would the rocket operate in outer space where there is no atmosphere? If so, how would you steer it? Could you brake it?

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