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An open jar of water moves in a vertical circle of radius \(0.50 \mathrm{~m}\) with a frequency that is small enough to put the water on the verge of falling out of the jar at the top of the circle. If the same demonstration were repeated on Mars, where the gravitational acceleration is only \(3.7 \mathrm{~m} / \mathrm{s}^{2}\), what is the change in the circling frequency to again put the water on the verge of falling out at the top point?

Short Answer

Expert verified
The frequency decreases by approximately 0.269 Hz on Mars.

Step by step solution

01

Understand the Problem

The problem asks to find the change in the frequency of the circular motion of a jar of water at the top point when moved to Mars, where the gravitational acceleration is different.
02

Formula for Critical Speed

To keep the water on the verge of falling out, the centripetal force equals the gravitational force at the top of the circle. The formula for critical speed at the top is \( v = \sqrt{g \cdot r} \), where \( g \) is the gravitational acceleration and \( r \) is the radius of the circle.
03

Calculate Critical Speed on Earth

First, calculate the critical speed using Earth's gravitational acceleration \( g = 9.8 \ m/s^2 \) and radius \( r = 0.5 \ m \). \[ v_{Earth} = \sqrt{9.8 \cdot 0.5} = \sqrt{4.9} \approx 2.21 \ m/s\]
04

Calculate Critical Speed on Mars

Next, calculate the critical speed using Mars's gravitational acceleration \( g = 3.7 \ m/s^2 \) and the same radius.\[ v_{Mars} = \sqrt{3.7 \cdot 0.5} = \sqrt{1.85} \approx 1.36 \ m/s\]
05

Relationship Between Speed and Frequency

The speed of an object in circular motion is related to frequency by the formula \( v = 2\pi r f \), where \( f \) is the frequency. Thus, \( f = \frac{v}{2\pi r} \).
06

Calculate Frequency on Earth and Mars

Calculate the frequency using the critical speeds calculated for Earth and Mars. For Earth:\[ f_{Earth} = \frac{2.21}{2\pi \cdot 0.5} \approx 0.703 \ Hz\]For Mars:\[ f_{Mars} = \frac{1.36}{2\pi \cdot 0.5} \approx 0.434 \ Hz\]
07

Determine the Change in Frequency

Find the change in frequency by subtracting the Earth frequency from the Mars frequency:\[ \text{Change in Frequency} = f_{Mars} - f_{Earth} = 0.434 - 0.703 \approx -0.269 \ Hz\]

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Key Concepts

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

Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. Imagine swinging a bucket of water around in a circle; the centripetal force is what keeps the water from flying out. This force always points towards the center of the circle.
For any object in circular motion, centripetal force depends on three key factors:
  • Mass of the object
  • Speed of the object
  • Radius of the circle
Centripetal force can be calculated using the formula: - \( F_c = \frac{mv^2}{r} \), where - \( m \) is the mass, - \( v \) is the velocity, and - \( r \) is the radius.
This concept explains why objects moving fast or in smaller circles require more force to stay on their path.
Critical Speed
Critical speed is the minimum speed an object must have at the top of a circular path to complete the loop without falling. In our jar of water example, it’s the speed at which the water is just about to spill.
At the top of the loop, the centripetal force is equal to the gravitational force acting downwards. The formula for critical speed is:- \( v = \sqrt{g \cdot r} \), where - \( g \) is the gravitational acceleration, and - \( r \) is the radius of the circle.
Understanding critical speed is crucial in various applications, such as designing roller coasters and ensuring satellites remain in their orbits.
Gravitational Acceleration
Gravitational acceleration is the constant rate at which objects accelerate towards the center of a massive body, like Earth or Mars. It is crucial for calculating the critical speed and resulting dynamics of circular motion.
On Earth, gravitational acceleration is approximately \( 9.8 \ m/s^2 \). On Mars, it is only about \( 3.7 \ m/s^2 \).
The difference in gravitational acceleration between planets affects not only the weight of objects but also how they move in circular paths. That’s why the critical speed of a jar of water in a circular motion differs between Mars and Earth.
Frequency of Motion
Frequency of motion in a circular path refers to how often an object completes a full circle in a given time. It is measured in hertz (Hz), which represents "cycles per second."
Frequency is directly tied to the object's speed and the circumference of the circle, according to the formula:- \( f = \frac{v}{2\pi r} \), where - \( v \) is velocity, - \( r \) is radius, and - \( 2\pi r \) is the circle's circumference.
When calculating frequency changes, like moving from Earth to Mars, it's essential to consider changes in gravitational acceleration, which lead to changes in critical speed and, consequently, frequency. In the exercise, we saw the frequency drop when transitioning from Earth to Mars due to lower gravitational acceleration.

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Most popular questions from this chapter

A flywheel turns through 40 rev as it slows from an angular speed of \(1.5 \mathrm{rad} / \mathrm{s}\) to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?

A disk, with a radius of \(0.25 \mathrm{~m}\), is to be rotated like a merrygo- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first 400 rad and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

A gyroscope flywheel of radius \(2.62 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min}\). (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

A wheel with a rotational inertia of \(0.50 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis is initially rotating at an angular speed of \(15 \mathrm{rad} / \mathrm{s}\). At time \(t=0\), a man begins to slow it at a uniform rate until it stops at \(t=\) \(5.0 \mathrm{~s}\). (a) By time \(t=3.0 \mathrm{~s}\), how much work had the man done? (b) For the full \(5.0 \mathrm{~s}\), at what average rate did the man do work?

A uniform metal pole of height \(30.0 \mathrm{~m}\) and mass \(100 \mathrm{~kg}\) is ini tially standing upright but then falls over to one side without it: lower end sliding or losing contact with the ground. What is the lin ear speed of the pole's upper end just before impact?

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