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\(\mathrm{A} 50.0 \mathrm{~kg}\) stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(95.0 \mathrm{~m} / \mathrm{s},\) what is the minimum radius of the circle so that the acceleration at this point will not exceed \(4.00 g ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Short Answer

Expert verified
The minimum radius of the circle such that the acceleration does not exceed 4g is 230 m. The apparent weight of the pilot at the lowest point of the pullout is 2450 N.

Step by step solution

01

Calculate the centripetal acceleration

We first calculate the maximum centripetal acceleration using the equation \(a = 4g\) where \(g = 9.8 \, m/s^2\). Hence the maximum centripetal acceleration is \(4 \times 9.8 = 39.2\, m/s^2\).
02

Find the minimum radius

We can now find the minimum radius where this acceleration will not exceed \(4g\). We can rearrange the equation for centripetal acceleration \(a = \frac{{v^2}}{{r}}\) to solve for \(r\): \(r = \frac{{v^2}}{{a}}\). Substituting the values \(v = 95\, m/s\) and \(a = 39.2\, m/s^2\) we find that \(r = \frac{{(95^2)}}{{39.2}} = 230\, m\).
03

Calculate the net force

Now find the net force at the bottom of the circular path, which is the sum of the force due to gravity and centripetal force \(F = m(a+g)\). Substituting with known values for \(m = 50\, kg\), \(a = 39.2\, m/s^2\) and \(g = 9.8\, m/s^2\), calculation results in \(F = 50 \times (39.2 + 9.8) = 2450\, N\).
04

Calculate apparent weight

Apparent weight is the reading that would be made by a scale spring between the pilot and the seat and equals the net force. Thus it equals 2450 N.

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Key Concepts

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

Circular Motion
Circular motion occurs when an object moves along a circular path. This can happen in various planes—horizontal, vertical, or even inclined.
If you imagine a plane performing a loop, it's essentially following a circular path. This path is defined by a continuous change in direction.

To keep moving in a circle, an object needs a continuously acting force directed towards the center of the circle. This is called centripetal force. It doesn't exist on its own; rather, it arises from other forces like tension, gravity, or friction.
  • In the stunt pilot's case, centripetal force comes from the gravitational pull acting in tandem with the speed of the plane.
  • This combination keeps the plane flying around the circular path.
Understanding how circular motion works is crucial for deciphering why certain forces act the way they do at specific points in the motion.
The same principles apply whether you are spinning a toy around on a string, or an airplane looping the loop. The faster the object moves for a given circle, the stronger the centripetal force required to keep the path consistent.
Apparent Weight
Apparent weight is what you "feel" as your weight in a non-standard environment, like in an accelerating car or a looping airplane. It's different from actual weight, which stems from gravitational force alone.

When the stunt pilot is at the lowest point in her dive's circular path, she feels a weight greater than her normal body weight.
This is because, at this point, the forces needed to change her plane's direction in the loop add up with gravity, making her feel "heavier."
  • In usual scenarios on flat ground, apparent and actual weight are the same.
  • However, during circular motion, they can differ wildly.
Imagine standing on a scale inside the plane; it doesn't just show your mass times gravitational acceleration (9.8 m/s^2). It adds the centripetal acceleration component due to circular motion.
The final reading, or apparent weight, is influenced by how fast you are moving and how tightly curved the path of motion is.
Net Force
Net force is the vector sum of all forces acting on an object. It's the total force that decides how an object moves according to Newton's Second Law of Motion.

In the context of circular motion, especially in vertical loops, calculating net force helps us understand the motion intricacies.
  • For the stunt pilot, the net force at the bottom of the loop combines the centripetal force—required to keep circular motion constant—and the force of gravity.
  • This means any force reading is actually reflecting more than just gravitational pull.
To calculate net force at the lowest point, the gravitational force (pilot's weight) and the force needed to keep moving in a circle are added together. This results in a net force larger than what the pilot would experience on a flat surface.
These calculations not only aid in understanding the forces at play but ensure safe and predictable aircraft maneuvers.

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Most popular questions from this chapter

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

Double Atwood's Machine. In Fig. \(\mathbf{P 5 . 1 1 4}\) masses \(m_{1}\) and \(m_{2}\) are connected by a light string \(A\) over a light, frictionless pulley \(B .\) The axle of pulley \(B\) is connected by a light string \(C\) over a light, frictionless pulley \(D\) to a mass \(m_{3} .\) Pulley \(D\) is suspended from the ceiling by an attachment to its axle. The system is released from rest. In terms of \(m_{1}, m_{2}, m_{3},\) and \(g,\) what are (a) the acceleration of block \(m_{3}\) (b) the acceleration of pulley \(B ;\) (c) the acceleration of block \(m_{1}\) (d) the acceleration of block \(m_{2} ;\) (e) the tension in string \(A ;\) (f) the tension in string \(C ?\) (g) What do your expressions give for the special case of \(m_{1}=m_{2}\) and \(m_{3}=m_{1}+m_{2} ?\) Is this reasonable?

A light rope is attached to a block with mass \(4.00 \mathrm{~kg}\) that rests on a friction less, horizontal surface. The horizontal rope passes over a friction less, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is \(15.0 \mathrm{~N}\). (a) Draw two free-body diagrams: one for each block. (b) What is the acceleration of either block? (c) Find \(m\). (d) How does the tension compare to the weight of the hanging block?

An \(8.00 \mathrm{~kg}\) block of ice, released from rest at the top of a 1.50-m-long friction less ramp, slides downhill, reaching a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of \(10.0 \mathrm{~N}\) parallel to the surface of the ramp?

An \(8.00 \mathrm{~kg}\) box sits on a ramp that is inclined at \(33.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is \(\mu_{\mathrm{k}}=0.300 .\) A constant horizontal force \(F=26.0 \mathrm{~N}\) is applied to the box (Fig. \(\mathbf{P} 5.73),\) and the box moves down the ramp. If the box is initially at rest, what is its speed \(2.00 \mathrm{~s}\) after the force is applied?

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