/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 A physics major is working to pa... [FREE SOLUTION] | 91Ó°ÊÓ

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A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius \(13.0 \mathrm{~m}\). She has mass \(70.0 \mathrm{~kg},\) and her motorcycle has mass \(40.0 \mathrm{~kg} .\) (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

Short Answer

Expert verified
The minimum speed at the top of the circle needs to be approximately 11.3 m/s to remain in contact with the sphere. The normal force exerted on the motorcycle at the bottom of the circle is approximately 8.4 kN.

Step by step solution

01

Compute the minimum speed to retain contact

Firstly, consider the situation at the top of the circle. For the motorcycle to stay in contact with the sphere, the gravitational force must be equal or less than the required centripetal force. Equating these two gives \(mg = \frac{mv^2}{r}\), where \(m\) is the total mass (mass of the motorcycle and the rider), \(g\) is the acceleration due to gravity, \(v\) is the velocity, and \(r\) is the radius of the circle. Solving for \(v\), the minimum speed comes out to be \(v = \sqrt{rg}\)
02

Calculate the value with given inputs

Substitute the values, \(m=110\) kg, \(r=13\) m, \(g=9.8\) m/s² into the equation, \(v = \sqrt{rg}\), to obtain the minimum speed necessary for the motorcycle to stay in contact with the sphere.
03

Compute the necessary normal force

Next, consider the situation at the bottom of the circle. The normal force and gravitational force are acting in the same direction (towards the center of the circle). So, the net force acting on the motorcycle is the sum of the gravitational force and the normal force, which should be equal to the centripetal force. This gives, \(mg + N = \frac{mv^2}{r}\), where \(N\) is the normal force. We know that at this point, \(v\) is twice the value calculated in the previous step.
04

Calculate the normal force

Solve the equation \(mg + N = \frac{mv^2}{r}\) for \(N\). This will give, \(N=\frac{mv^2}{r} - mg\). Substituting the values calculated earlier, we can find the magnitude of the normal force exerted on the motorcycle by the sphere at the bottom of the circle.

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Key Concepts

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

Centripetal Force
When a motorcycle or any body is moving in a circle, it experiences a force called centripetal force. This force is necessary to make the object follow a circular path. It acts towards the center of the circle and is responsible for changing the direction of the object's velocity, keeping it on the circular path.

In the context of our motorcycle scenario, the centripetal force is vital for the rider to stay on that circular track inside the sphere. It's calculated using the formula:
  • \( F_c = \frac{mv^2}{r} \)
Here, \( m \) is the total mass moving in the circle, \( v \) is the speed or velocity of the object, and \( r \) is the radius of the circle.

For the motorcycle to maintain contact while traveling in the vertical circle, the necessary centripetal force at the top of the circle must at least equal the gravitational pull acting on the motorcycle and rider combined.
Gravitational Force
Gravitational force is the force exerted by the Earth that pulls objects towards its center. It is what gives weight to physical objects and is calculated as:
  • \( F_g = mg \)
Where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on the surface of the Earth.

In the vertical circle example, at the top of the loop, the gravitational force is acting downwards. For the motorcycle to remain in contact with the sphere, this force needs to be accounted for as it contributes to the total centripetal force required. Essentially, at the top of the circle, the gravitational force helps to provide the necessary centripetal force, and if the speed is appropriate, these forces will balance properly.
Normal Force
Normal force acts at a perpendicular angle from the contact surface. It is the force that prevents objects from "falling" through the surface they rest on. When traveling in a circle, especially at high speeds, the normal force plays a crucial role in determining how much pressure an object experiences.

In the given problem, at the bottom of the circle, the normal force works along with the gravitational force to ensure the motorcycle stays on its path inside the sphere. The normal force can be calculated from the equation:
  • \( N = \frac{mv^2}{r} - mg \)
Here, this equation acknowledges the extra speed the motorcycle has gained, given that its velocity at the bottom of the circle is twice the speed needed at the top.

Normal force increases as speed increases, which explains why the rider feels pushed against the sphere at the bottom of the circle.
Minimum Speed
To maintain contact at the top of a vertical circle, an object must reach a certain minimum speed. This speed ensures that the force pulling it towards the center of the circle -- the centripetal force -- is present to prevent it from falling.

The formula for calculating this minimum speed is derived by setting the gravitational force equal to the centripetal force needed at the top of the circle:
  • \( v = \sqrt{rg} \)
Where \( r \) is the radius of the circle and \( g \) is the acceleration due to gravity.

Achieving this speed means that the motorcycle won't lose contact with the surface of the sphere due to the pull of gravity being balanced by the inertia of the speed in a circular path.
Vertical Circle
A vertical circle involves an object moving around a circular path while changing its height relative to Earth's surface. Gravitational forces play a significant role here, as they not only influence the centripetal force needed but can also help provide it when the object is at the circle's peak.

For an object like a motorcycle inside a sphere to perform a vertical circle, it needs to maintain sufficient speed to overcome gravitational forces at the top. At the same time, it needs to manage the increased forces at the bottom where both the gravitational and normal forces add up.

Understanding motion in a vertical circle helps one grasp how different forces and speeds interact to allow continuous motion in such a loop, much like what amusement park rides and certain sports applications utilize.

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Most popular questions from this chapter

A \(2540 \mathrm{~kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t)=A t+B t^{2}\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) at the instant of ignition and, 1.00 s later, an upward velocity of \(2.00 \mathrm{~m} / \mathrm{s}\). (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

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?

A small rock with mass \(m\) is attached to a light string of length \(L\) and whirled in a vertical circle of radius \(R\). (a) What is the minimum speed \(v\) at the rock's highest point for which it stays in a circular path? (b) If the speed at the rock's lowest point in its circular path is twice the value found in part (a), what is the tension in the string when the rock is at this point?

Runway Design. A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is \(700 \mathrm{~kg}\), and the total resistance (air drag plus friction with the runway on each may be assumed constant and equal to \(2500 \mathrm{~N}\). The tension in the towrope between the transport plane and the first glider is not to exceed \(12,000 \mathrm{~N}\). (a) If a speed of \(40 \mathrm{~m} / \mathrm{s}\) is required for takeoff, what minimum length of runway is needed? (b) What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?

Forces During Chin-ups. When you do a chin-up, you raise your chin just over a bar (the chinning bar), supporting yourself with only your arms. Typically, the body below the arms is raised by about \(30 \mathrm{~cm}\) in a time of \(1.0 \mathrm{~s},\) starting from rest. Assume that the entire body of a \(680 \mathrm{~N}\) person doing chin-ups is raised by \(30 \mathrm{~cm},\) and that half the \(1.0 \mathrm{~s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Draw a free-body diagram of the person's body, and use it to find the force his arms must exert on him during the accelerating part of the chin-up.

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