/*! 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 118 A physics major is working to pa... [FREE SOLUTION] | 91Ó°ÊÓ

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A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m} .\) The physics major has mass \(70.0 \mathrm{kg},\) and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere? (b) At the bottom of the circle, his 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
(a) Minimum speed at top: 11.29 m/s. (b) Normal force at bottom: 5374 N.

Step by step solution

01

Identify the Forces at the Top of the Circle

At the top of the circle, the forces acting on the physics major and motorcycle system are gravity and the normal force. For the motorcycle to maintain contact, the centripetal force is equal to the gravitational force plus the normal force. However, minimum speed implies that the normal force approaches zero.
02

Calculate Minimum Speed at the Top

To find the minimum speed required at the top of the circle, set the gravitational force equal to the centripetal force. The formula for gravitational force is \( F_g = (m_{rider} + m_{cycle})g \) where \( g \) is 9.8 \( m/s^2 \). The formula for centripetal force is \( F_c = \frac{(m_{rider} + m_{cycle})v^2}{r} \). Setting these equal gives \( (m_{rider} + m_{cycle})g = \frac{(m_{rider} + m_{cycle})v^2}{r} \). Cancel out the masses and solve for \( v \): \( v = \sqrt{gr} \).
03

Substitute Values for Minimum Speed

Substitute the known values for \( g \) and \( r \): \( v = \sqrt{9.8 \times 13.0} \). Calculating these gives the minimum speed \( v = \sqrt{127.4} \approx 11.29 \ m/s \).
04

Calculate Speed at the Bottom of the Circle

The speed at the bottom of the circle is twice the speed at the top, so use the previous result: \( v_{bottom} = 2 \times 11.29 \approx 22.58 \ m/s \).
05

Determine Normal Force at the Bottom

At the bottom of the circle, the centripetal force is provided by the net force, which is the normal force minus the gravitational force. The equation is \( N - (m_{rider} + m_{cycle})g = \frac{(m_{rider} + m_{cycle})v^2}{r} \). Solve for the normal force \( N = (m_{rider} + m_{cycle})g + \frac{(m_{rider} + m_{cycle})v^2}{r} \).
06

Substitute Values for Normal Force Calculation

Substitute the known values into the equation: \( N = (70 + 40) \times 9.8 + \frac{(70 + 40) \times (22.58)^2}{13} \). Calculating these gives \( N \approx 1078 + 4296 \approx 5374 \ N \).

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Key Concepts

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

Normal Force
Normal force is the force exerted by a surface to support the weight of an object resting on it. It's perpendicular to the surface contact. In a scenario where the physics major rides a motorcycle inside a sphere, the concept of normal force becomes crucial, especially at the bottom of the circle. In such a case, the normal force helps maintain the circular motion by providing the necessary centripetal force.

In vertical circular motion, the magnitude of the normal force varies depending on the position. At the top of the circle, the force may become very small, close to zero, indicative of reaching the minimum speed necessary to maintain contact. At this point, it is the gravitational force that ensures the motorcycle stays on the sphere.

Conversely, at the bottom of the circle, the normal force is much greater because it has to compensate for both gravity and the centripetal force needed for maintaining circular motion. This is why calculating the normal force, especially at this point, provides insights into the necessary strength and safety for activities like motorbike riding in a sphere.
Gravitational Force
Gravitational force is the attractive force exerted by anything that has mass. On Earth, it is commonly represented as the weight of an object, calculated by mass multiplied with the gravitational acceleration, 9.8 m/s².

In this exercise, gravitational force plays a pivotal role at the top of the circle. It's the force that keeps the motorcycle in contact with the sphere when the normal force is minimized. Using the formula:
  • The force due to gravity is given by: \( F_g = (m_{rider} + m_{cycle})g \).
This gravitational force provides the necessary centripetal acceleration acting downward. This means, at minimum speed, the gravitational force alone is responsible for the motion of the motorcycle, maintaining its path around the circle.

The interplay of gravitational and normal forces illustrates how these fundamental forces ensure movement in a closed circular path, underscoring the principles of physics that govern carnival motorbike stunts.
Minimum Speed Calculation
The minimum speed calculation is essential to keep objects in constant contact with the surface of a circular path. To maintain contact at the loop's topmost point, the centripetal force must at least equal the gravitational force. This means any additional force, such as the normal force, becomes negligible.

The calculation involves equating the centripetal force, dependent on velocity squared, to the gravitational force. By simplifying the formula, you derive:
  • The minimum speed is expressed as: \( v = \sqrt{gr} \).
Substituting the values of gravitational acceleration \((g = 9.8 \ m/s^2)\) and the circle's radius \((r = 13 \ m)\), you compute:
  • \( v = \sqrt{9.8 \times 13} \approx 11.29 \ m/s \).
This calculated speed ensures that, at the top of the loop, there is enough velocity to maintain circular motion without losing contact. The minimum speed acts as a benchmark for safety and skill required in such circus physics feats.

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Most popular questions from this chapter

A 8.00 -kg block of ice, released from rest at the top of a \(1.50-\mathrm{m}\)-long frictionless 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?

Block \(A\) in Fig. P5.72 weighs 60.0 \(\mathrm{N} .\) The coefficient of static friction between the block and the surface on which it rests is \(0.25 .\) The weight \(w\) is 12.0 \(\mathrm{N}\) and the system is in equilibrium. (a) Find the friction force exerted on block \(A .\) (b) Find the maximum weight \(w\) for which the system will remain in equilibrium.

A machine part consists of a thin 40.0 -cm-long bar with small 1.15 -kg masses fastened by screws to its ends. The screws can support a maximum force of 75.0 \(\mathrm{N}\) without pulling out. This bar rotates about an axis perpendicular to it at its center. (a) As the bar is turning at a constant rate on a horizontal, frictionless surface, what is the maximum speed the masses can have without pulling out the screws? (b) Suppose the machine is redesigned so that the bar turns at a constant rate in a vertical circle. Will one of the screws be more likely to pull out when the mass is at the top of the circle or at the bottom? Use a free-body diagram to see why. (c) Using the result of part (b), what is the greatest speed the masses can have without pulling a screw?

A 125 -kg (including all the contents) rocket has an engine that produces a constant vertical force (the thrust) of 1720 \(\mathrm{N}\) . Inside this rocket, a 15.5 -N electrical power supply rests on the floor. (a) Find the acceleration of the rocket. (b) When it has reached an altitude of \(120 \mathrm{m},\) how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

A 12.0 -kg box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{\mathrm{s}}=0.19\) and \(\mu_{\mathrm{k}}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

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