/*! 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 54 A student builds and calibrates ... [FREE SOLUTION] | 91Ó°ÊÓ

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A student builds and calibrates an accelerometer, which she uses to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with her observes that the plumb bob hangs at an angle of \(15.0^{\circ}\) from the vertical when the car has a speed of \(23.0 \mathrm{m} / \mathrm{s} .\) (a) What is the centripetal acceleration of the car rounding the curve? (b) What is the radius of the curve? (c) What is the speed of the car if the plumb bob deflection is \(9.00^{\circ}\) while round. ing the same curve?

Short Answer

Expert verified
The centripetal acceleration of the car rounding the curve is \(2.57 \mathrm{m}/\mathrm{s}^2\). The radius of the curve is \(205.71 \mathrm{m}\). The speed of the car when the plumb bob is making a \(9.0^{\circ}\) deflection angle is \(19.00 \mathrm{m}/\mathrm{s}\).

Step by step solution

01

Calculating the Centripetal Acceleration

We know that the centripetal force is balanced by the gravity and the force acting perpendicular to the gravity. Hence, the component of gravity in the horizontal direction is equal to the centripetal force. The magnitude of gravitational acceleration acting downwards (g) is \(9.8 \mathrm{m}/\mathrm{s}^2\). Calculate the centripetal acceleration using the relation: \( a_c = g \cdot tan(θ) \), where θ is \(15.0^{\circ}\). This will give: \( a_c = 9.8 \cdot tan(15.0^{\circ}) \)
02

Calculating the Radius of the Curve

The formula for centripetal acceleration when a body moves in a circle of radius r with speed v is \(a_c = v^2/r\). Rearrange the formula for r, and plug in the values of a_c calculated in step 1, and v = \(23.0 \mathrm{m}/\mathrm{s}\). The formula to calculate r will be: \( r = v^2/a_c \)
03

Calculating the New Speed of the Car

Now we need to calculate the new speed of the car when the deflection angle is \(9.0^{\circ}\). We use the same relations established in steps 1 and 2. Firstly, find the new centripetal acceleration \(a_c' = g \cdot tan(9.0^{\circ})\). As the radius of the curve remains the same, we can use the relation \(v' = \sqrt{a_c' \cdot r}\) to find the new speed of the car.

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Key Concepts

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

Acceleration in Circular Motion
When an object moves in a circular path, it experiences a type of acceleration known as centripetal acceleration. This is not due to an increase in the speed of the object, but rather a change in its direction. Imagine swinging a ball attached to a string in a circular path; your hand provides a constant force towards the center, keeping the ball moving in that circle. This inward force is necessary for any kind of circular motion.

Centripetal acceleration, denoted as \(a_c\), always points towards the center of the circle and can be calculated with the formula \(a_c = \frac{v^2}{r}\), where \(v\) is the velocity of the object and \(r\) is the radius of the circular path. A greater speed or a smaller radius will result in a larger centripetal acceleration.
Radius of Curvature
The radius of curvature refers to the radius of the imaginary circle that best describes the path an object in motion follows when it takes on a curved trajectory. This concept is pivotal when studying circular motion because it helps in understanding the tightness or sharpness of a turn.In the context of our exercise, the radius of curvature of the car's path can be determined using the centripetal acceleration formula rearranged to solve for the radius, \(r = \frac{v^2}{a_c}\). A sharper turn - that is, a smaller radius - necessitates a greater centripetal force to maintain circular motion at a constant speed. Conversely, a broader curve, with a larger radius, requires less centripetal force. This relationship highlights how the curvature of the car’s path is intrinsically linked to the centripetal force acting upon it.
Plumb Bob Accelerometer
A plumb bob accelerometer is an ingenious yet straightforward device used to measure acceleration. The device consists of a weight, or plumb bob, suspended from a fixed point, typically by a string, and a protractor to measure the angle of displacement from the vertical. In a state of rest or uniform motion, the plumb bob hangs vertically, indicating no acceleration. However, when acceleration occurs, inertia causes the plumb bob to deflect at an angle proportional to the acceleration.In our student's experiment, the angle of deflection measured by the protractor corresponds to the lateral or centripetal acceleration the car experiences while maneuvering the curve. Using the tilt angle, one can calculate centripetal acceleration through the formula \(a_c = g \cdot \tan(\theta)\), where \(g\) is the acceleration due to gravity and \(\theta\) is the measured angle. This type of accelerometer is simple but effective for demonstrating basic principles of physical accelerations.

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Most popular questions from this chapter

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of \(0.0337 \mathrm{m} / \mathrm{s}^{2},\) while a point at the poles experiences no centripetal acceleration. (a) Show that at the equator the gravitational force on an object must exceed the normal force required to support the object. That is, show that the object's true weight exceeds its apparent weight. (b) What is the apparent weight at the equator and at the poles of a person having a mass of \(75.0 \mathrm{kg} ?\) (Assume the Earth is a uniform sphere and take \(g=9.800 \mathrm{m} / \mathrm{s}^{2} .\) )

A 0.142-kg baseball has a terminal speed of \(42.5 \mathrm{m} / \mathrm{s}\) \((95 \mathrm{mi} / \mathrm{h}) \cdot\) (a) If a baseball experiences a drag force of magnitude \(R=C v^{2},\) what is the value of the constant \(C ?\) (b) What is the magnitude of the drag force when the speed of the baseball is \(36.0 \mathrm{m} / \mathrm{s} ?\) (c) Use a computer to determine the motion of a bascball thrown vertically upward at an initial speed of \(36 \mathrm{m} / \mathrm{s} .\) What maximum height does the ball reach? How long is it in the air? What is its speed just before it hits the ground?

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of \(20.0^{\circ}\) with the horizontal. A piece of luggage having mass \(30.0 \mathrm{kg}\) is placed on the carousel, \(7.46 \mathrm{m}\) from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, \(7.94 \mathrm{m}\) from the axis of rotation. Now going around once in every \(34.0 \mathrm{s}\), the bag is on the verge of slipping. Calculate the coefficient of static friction between the bag and the carousel.

A \(40.0-\mathrm{kg}\) child swings in a swing supported by two chains, each \(3.00 \mathrm{m}\) long. If the tension in each chain at the lowest point is \(350 \mathrm{N},\) find \((\mathrm{a})\) the child's speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Neglect the mass of the seat.)

A small container of water is placed on a carousel inside a microwave oven, at a radius of \(12.0 \mathrm{cm}\) from the center. The turntable rotates steadily, turning through one revolution in each 7.25 s. What angle does the water surface make with the horizontal?

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