/*! 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 24 A small container of water is pl... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
The angle the water surface makes with the horizontal is approximately \(43.8 \mathrm{degrees}\).

Step by step solution

01

Identify Known Values

We are given that the radius \(r\) is \(12.0 \mathrm{cm}\) and the time taken to complete one revolution (\(T\)) is \(7.25 \mathrm{s}\).
02

Calculate the Rotational Speed

Rotational speed (\(\omega\)) can be calculated using the formula \(\omega = 2\pi / T\). Substituting the given time (\(T = 7.25 \mathrm{s}\)), we get \(\omega = 0.867 \mathrm{rad/s}\).
03

Find the Centripetal Acceleration

Centripetal acceleration (a) equals the square of the rotational speed times the radius (a = \(r\omega^2\)). Substituting the known values into the formula, we get centripetal acceleration \(\approx 9.4 \mathrm{m/s^2}\).\n
04

Calculate the Angle

The angle (\(\theta\)) the water surface makes with the horizontal can be calculated by \(\theta = \tan^{-1}(a/g)\), where \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)). Substituting the calculated acceleration into this formula gives \(\theta \approx 43.8 \mathrm{degrees}\).

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Key Concepts

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

Rotational Motion
Imagine a merry-go-round spinning at your local fair. The horses and seats are fixed on a platform that rotates around a central point — this is one of the simplest examples of rotational motion in action. Rotational motion refers to the movement of an object in a circular path around a fixed axis. In our textbook exercise, we see rotational motion demonstrated by a carousel inside a microwave oven.

It's important to realize that every point on the rotating carousel follows a circular trajectory. The path each point takes is determined by its distance from the center, or radius. Think of the outer horses on the merry-go-round — they cover a larger distance per rotation and feel a stronger push outward compared to those closer to the center. This is due to rotational motion's influence on the 'centripetal force', which we'll discuss in more detail later.
Angular Velocity
Angular velocity is a fancy term for how fast something is spinning or rotating. It tells us the rate at which the angle is changing as an object follows a circular path. The unit for measuring angular velocity is radians per second (rad/s).

In the context of our exercise problem — a carousel microwave's turntable — we calculate the angular velocity by taking the full rotation in radians (which is always \(2\pi\), since a full circle is \(360\degree\) or \(2\pi\) radians) and divide it by the time, \(T\), it takes to complete one revolution. If a carousel takes 7.25 seconds for a complete turn, its angular velocity is the rotation (\(2\pi\) radians) divided by 7.25 seconds, giving us the speed of rotation in radians per second.
Centripetal Force
Centripetal force is not a type of force like gravity or friction. Instead, it's really just the name we give to whatever force is keeping an object moving in a circle, pointing towards the center of the rotation. Without this inward force, objects would fly off in a straight line due to inertia! Astronauts in training feel this force when they're spun around in a centrifuge — it's what pushes them against the wall.

In the microwave oven example, water inside the carousel experiences centripetal force. This force keeps the water from spilling as the turntable rotates. We calculate the magnitude of this force for any object moving in a circular path with the formula \(F_c = m \times (r \times \(\omega^2\))\) where \(m\) is the mass of the object, \(r\) the radius, and \(\omega\) the angular velocity.
Acceleration Due to Gravity
Imagine dropping a ball from the top of a building. This ball speeds up as it falls, correct? This acceleration, or 'speeding up,' happens because of Earth's gravity. We call this 'acceleration due to gravity', symbolized as \(g\), and it has a value of approximately \(9.81 \mathrm{m/s^2}\) on the surface of the Earth.

Accurately factoring in the acceleration due to gravity is crucial when calculating the angle the water surface makes with the horizontal in a rotating system, like our microwave carousel example. Here, we have to compare the outward push due to the centripetal force (remember, it's the 'apparent' force pushing outward due to rotation, not an actual force!) with the real force of gravity pulling down. By doing this, we can determine the tilt of the water's surface, generally using a bit of trigonometry as seen in the angle calculation from our textbook exercise.

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Most popular questions from this chapter

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time. The equation \(v=v_{i}+a t\) gives its velocity \(v\) as a function of time, where \(a\) is its constant acceleration. What if velocity is instead a linear function of position? Assume that as a particular object moves through a resistive medium, its speed decreases as described by the equation \(v=v_{i}-k x,\) where \(k\) is a constant coefficient and \(x\) is the position of the object. Find the law describing the total force acting on this object.

In a cyclotron (one type of particle accelerator), a deuteron (of atomic mass \(2.00 \mathrm{u}\) ) reaches a final speed of \(10.0 \%\) of the speed of light while moving in a circular path of radius \(0.480 \mathrm{m}\). The deuteron is maintained in the circular path by a magnetic force. What magnitude of force is required?\(214.897 N\)

A \(0.400-\mathrm{kg}\) object is swung in a vertical circular path on a string \(0.500 \mathrm{m}\) long. If its speed is \(4.00 \mathrm{m} / \mathrm{s}\) at the top of the circle, what is the tension in the string there?

A \(50.0-\mathrm{kg}\) parachutist jumps from an airplane and falls to Earth with a drag force proportional to the square of the speed, \(R=C v^{2} .\) Take \(C=0.200 \mathrm{kg} / \mathrm{m}\) (with the parachute closed) and \(C=20.0 \mathrm{kg} / \mathrm{m}\) (with the chute open). (a) Determine the terminal speed of the parachutist in both configurations, before and after the chute is opened. (b) Set up a numerical analysis of the motion and compute the speed and position as functions of time, assuming the jumper begins the descent at \(1000 \mathrm{m}\) above the ground and is in free fall for 10.0 s before opening the parachute. (Suggestion: When the parachute opens, a sudden large acceleration takes place; a smaller time step may be necessary in this region.)

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?

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