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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the circle's radius is \(10 \mathrm{~m}\). At the top of the circle, what are the (a) magnitude \(F_{B}\) and (b) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
At 5 m/s, \( F_B = -3724.5 \, \text{N} \) downward; at 12 m/s, \( F_B = 2348.8 \, \text{N} \) upward.

Step by step solution

01

Determine the gravitational force

The weight of the car and riders is given as 5000 N. This weight acts downward at every point in the circle. The gravitational force is essentially the weight of the object, so in this case, \( F_g = 5000 \, \text{N} \).
02

Use centripetal force formula

At the top of the circle, the centripetal force needed is provided by the sum of the gravitational force and the force provided by the boom, which we label as \( F_B \). The formula for centripetal force is \( F_c = \frac{mv^2}{r} \). First, we need to find the mass \( m \). Since weight \( W = mg = 5000 \, \text{N} \), so \( m = \frac{5000}{9.8} = 510.2 \, \text{kg} \).
03

Calculate the centripetal force for \( v = 5 \, \text{m/s} \)

Substitute the mass \( m = 510.2 \, \text{kg} \), speed \( v = 5 \, \text{m/s} \), and radius \( r = 10 \, \text{m} \) into the centripetal force formula: \[ F_c = \frac{510.2 \times 5^2}{10} = 1275.5 \, \text{N}. \]
04

Determine \( F_B \) when \( v = 5 \, \text{m/s} \)

At the top of the trajectory, the gravitational force and the force from the boom both contribute to the centripetal force, so \( F_B = F_c - F_g = 1275.5 - 5000 \). Since \( F_B \) is negative, this indicates that the force from the boom must also act downwards to reduce the required support: \( F_B = -3724.5 \, \text{N} \).
05

Calculate the centripetal force for \( v = 12 \, \text{m/s} \)

Substitute the new speed \( v = 12 \, \text{m/s} \) into the centripetal force formula: \[ F_c = \frac{510.2 \times 12^2}{10} = 7348.8 \, \text{N}. \]
06

Determine \( F_B \) when \( v = 12 \, \text{m/s} \)

With this new speed, using the same principle as before, \( F_B = F_c - F_g = 7348.8 - 5000 = 2348.8 \, \text{N}. \) Since \( F_B \) is positive, it suggests that the boom must provide additional upward force against gravity.

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental force responsible for the attraction between masses. It acts downwards toward the center of the Earth. In the context of the amusement park ride, the gravitational force is synonymous with the weight of the car and riders, which was given as 5000 Newtons. This weight is constant and affects the motion at every point along the vertical circle.

Gravitational force is calculated using the formula: \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth).
  • At the top of the circle, gravitational force plays a crucial role as it combines with other forces either to assist or to be countered depending on the speed of the car.
  • Understanding gravitational force helps in analyzing the forces at play when an object is in vertical circular motion.
Centripetal Acceleration
Centripetal acceleration is vital for any object moving in a circle. It points towards the center of the circle, maintaining the object's curved path. The formula for centripetal acceleration is \( a_c = \frac{v^2}{r} \), where \( v \) is the speed of the object and \( r \) is the radius of the circle.

In the case of the amusement park ride, it's essential to understand that centripetal acceleration enables the car to stay on its circular path.
  • The centripetal acceleration depends on both the speed of the ride and the radius of the circular path.
  • As speed increases, more force is required to maintain the circular motion, equating to greater centripetal acceleration.

For instance, when the car is at a speed of 5 m/s, its centripetal acceleration is significantly lower as compared to when the speed reaches 12 m/s. This change in speed requires the forces acting on the car to adjust accordingly to sustain its motion in a circle.
Vertical Circular Motion
Vertical circular motion is a type of motion where an object travels in a circular path vertically. This can be seen in roller coaster loops or rides similar to the amusement park ride discussed here. The critical aspect of this motion is the combination of gravitational force and the force necessary to keep the object on its path, essentially the centripetal force.

At different points in a vertical circle, forces interact uniquely due to gravity's constant pull downwards.
  • At the top of the circle, as seen with the ride, both gravity and the force from the boom can act downward which decreases the need for additional force from the boom if the speed is lower.
  • Conversely, at higher speeds, the force needed from the boom increases to counteract gravity and maintain the circular motion.

This dual-force requirement explains why the calculations for \( F_B \) show a change in direction with different speeds. A deeper understanding of vertical circular motion and these concepts helps explain why rides feel different at various speeds and points along the loop.

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Most popular questions from this chapter

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