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A model airplane with mass \(0.750 \mathrm{~kg}\) is tethered by a wire so that it flies in a circle \(30.0 \mathrm{~m}\) in radius. The airplane engine provides a net thrust of \(0.800 \mathrm{~N}\) perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to its flight path.

Short Answer

Expert verified
The net thrust produces a torque of \(24.0 N.m\), the airplane has an angular acceleration of \(0.0356 rad/s^2\), and a linear acceleration of \(1.07 m/s^2\).

Step by step solution

01

Find the torque produced about the center.

Torque (\(\tau\)) is given by the formula \(\tau = rF\sin(\theta)\), where \(r\) is the radius, \(F\) is the force, and \(\theta\) is the angle between the r and F. Here, the force is the net thrust and the angle is 90 degrees since the force is perpendicular to the tethering wire. Plug in the given values: radius \(r = 30.0 \ m\), net thrust \(F = 0.800 \ N\), and angle \(\(\theta = 90 degrees\). \[\tau = 30.0 m * 0.800 N * \sin(90°)\]\[\tau = 24.0 N.m\]
02

Find the angular acceleration of the airplane.

Angular acceleration (\(\alpha\)) is given by the formula \(\alpha = \tau/I\), where \(\tau\) is the torque and \(I\) is the moment of inertia. The moment of inertia for an object in circular motion is given by \(I = m*r^2\), where \(m\) is the mass and \(r\) is the radius. Here, \(m = 0.750 kg\) and \(r = 30.0 m\). Plug these values into the formula for \(I\): \[I = 0.750 kg * (30.0 m)^2 = 675 kg.m^2\] Now, plug in the values for \(\tau\) and \(I\) into the formula for \(\alpha\): \[\alpha = 24.0 N.m / 675 kg.m^2 = 0.0356 rad/s^2\]
03

Find the linear acceleration of the airplane.

The linear acceleration (\(a\)) of an object in circular motion is related to the angular acceleration (\(\alpha\)) by the equation \(a = r*\alpha\), where \(r\) is the radius. Here, r = 30.0 m and \(\alpha = 0.0356 rad/s^2\). Plug in these values: \[a = 30.0 m * 0.0356 rad/s^2 = 1.07 m/s^2\]

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Key Concepts

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

Torque in Physics
When studying circular motion, as in the case of a tethered airplane, the concept of torque is fundamental. Torque, represented by the symbol \(\tau\), is a measure of the force that causes an object to rotate about an axis. The larger the torque, the greater the object's tendency to spin.

Torque is calculated using the equation \(\tau = rF\sin(\theta)\), where \(r\) is the distance from the pivot point to the point where the force is applied, \(F\) is the magnitude of the force, and \(\theta\) is the angle between the direction of the force and the radius vector from the pivot point to the point of force application.

Let's apply this to our airplane example. The engine provides a thrust forming a right angle to the tether; thus, \(\sin(90°)\) equals 1. If the force applied is perpendicular (\(\theta=90°\)), the calculation simplifies to \(\tau = rF\), and no thrust component is lost due to angular misalignment. The computed torque of 24 N.m indicates that a considerable rotational force is applied to the airplane flying in a circle.
Angular Acceleration
Following the notion of torque is the concept of angular acceleration (\(\alpha\)). Angular acceleration tells us how the angular velocity of an object changes with time. It's comparable to linear acceleration but for rotating systems. The angular acceleration is given by \(\alpha = \tau/I\), where \(\tau\) is the torque and \(I\) is the moment of inertia, which measures an object's resistance to changes in its rotational motion.

The moment of inertia itself depends on the mass distribution of the object. For a point mass moving in a circle, as with the tethered airplane, it is calculated as \(I = m*r^2\). When the values for torque and the moment of inertia are substituted into the formula, we can see the direct proportionality between torque and angular acceleration, given a constant moment of inertia.

With a calculated angular acceleration of 0.0356 rad/s², our airplane will change its angular velocity by this amount every second it's in level flight, assuming no other forces are acting and the mass distribution does not change.
Linear Acceleration
Finally, we examine linear acceleration (\(a\)), which is the rate of change of an object's velocity in a straight line. While angular acceleration concerns rotation, linear acceleration deals with motion along a path. In circular motion scenarios, such as the airplane travelling around the circle's circumference, linear acceleration can be derived from angular acceleration using the conversion formula \(a = r*\alpha\).

This relationship shows that the linear acceleration of an object moving in a circle is proportional to both the radius of the circle and the angular acceleration. It signifies how much the velocity of the airplane changes along its circular path. With the calculated value of 1.07 m/s², the airplane not only rotates faster over time due to angular acceleration but also gains speed tangentially to its flight path.

Understanding the Relation Between Angular and Linear Acceleration

It's important to understand that angular and linear acceleration describe the same motion from different perspectives. In the tethered airplane exercise, while the angular acceleration measures how quickly the plane spins around the circle's center, the linear acceleration indicates how quickly its actual speed around the circuit changes.

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Most popular questions from this chapter

A \(10.0-\mathrm{kg}\) cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of \(10.0 \mathrm{~m} / \mathrm{s}\), determine (a) the translational kinetic energy of its center of gravity, (b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.

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