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Chapter 9: Problem 33

A mass of \(1.5\) kg out in space moves in a circle of radius \(25 \mathrm{~cm}\) at a constant \(2.0\) rev/s. Calculate \((a)\) the tangential speed, \((b)\) the acceleration, and ( \(c\) ) the required centripetal force for the motion.

Short Answer

Expert verified
(a) \( \pi \) m/s, (b) \( 39.48 \) m/s虏, (c) \( 59.22 \) N.

Step by step solution

01

Convert Units

First, convert the radius from centimeters to meters. Since 1 meter = 100 centimeters, therefore the radius in meters is \( r = 25 \text{ cm} = 0.25 \text{ m} \).
02

Calculate Tangential Speed

The tangential speed \( v \) can be calculated using the formula \( v = 2\pi r f \), where \( f \) is the frequency of rotation in revolutions per second. Here, \( f = 2.0 \mathrm{~rev/s} \), and substituting the known values, \( v = 2 \pi \times 0.25 \times 2 = \pi \) m/s.
03

Calculate Acceleration

The centripetal acceleration \( a_c \) is given by \( a_c = \frac{v^2}{r} \). Substituting the values from the previous step, \( a_c = \frac{\pi^2}{0.25} \approx 39.48 \text{ m/s}^2 \).
04

Calculate Centripetal Force

The centripetal force \( F_c \) required for circular motion is given by \( F_c = m a_c \), where \( m \) is the mass. Substituting the mass \( m = 1.5 \) kg and the centripetal acceleration \( a_c = 39.48 \text{ m/s}^2 \), we find \( F_c = 1.5 \times 39.48 = 59.22 \) N.

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Key Concepts

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

Tangential Speed
Tangential speed refers to how fast an object is moving along its path in circular motion. It is perpendicular to the radius of the circle and crucial for understanding how different rotational speeds translate into linear motion. To find tangential speed in circular motion, we often use the formula:
  • \( v = 2\pi r f \)
where \( v \) is the tangential speed, \( r \) is the radius of the circle, and \( f \) is the frequency of rotation.
In our exercise, the object has a frequency of 2 revolutions per second and a radius of 0.25 meters. This makes the tangential speed \( \pi \) m/s after plugging the values into the formula. Understanding tangential speed is important because it directly relates the rotational motion to linear speed, helping in applications like designing gear systems and understanding planetary orbits.
Centripetal Acceleration
Centripetal acceleration is necessary for an object to maintain circular motion. It acts towards the center of the circle and ensures the object stays on its path. The equation that helps us determine this acceleration is:
  • \( a_c = \frac{v^2}{r} \)
Here, \( a_c \) is the centripetal acceleration, \( v \) is the tangential speed, and \( r \) is the radius.
Using the tangential speed calculated earlier, we find that the centripetal acceleration is approximately 39.48 m/s虏.
This concept is crucial as it explains why objects in circular paths feel an "inward" force and provides insights into dynamics at play in rides like roller coasters or rotating planetary bodies.
Centripetal Force
Centripetal force is what keeps an object moving in a circular path and is directed towards the center of the circle. Without it, the object would move off in a straight line due to inertia. The centripetal force can be determined using the formula:
  • \( F_c = m a_c \)
where \( F_c \) is the centripetal force, \( m \) is the mass of the object, and \( a_c \) is the centripetal acceleration. In the situation given, with a mass of 1.5 kg and an acceleration of 39.48 m/s虏, the required centripetal force is calculated to be 59.22 Newtons.
Comprehending centripetal force is essential for the safe design of physical systems involving rotation, such as vehicles making turns, satellites orbiting Earth, or objects being whirled around on a string.

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Most popular questions from this chapter

Express \(40.0 \mathrm{deg} / \mathrm{s}\) in \((a) \mathrm{rev} / \mathrm{s},(b) \mathrm{rev} / \mathrm{min}\), and \((c) \mathrm{rad} / \mathrm{s}\).

A pebble is stuck in the tread of a tire having a diameter of \(80.0\) \(\mathrm{cm}\). The tire spins through \(23.5\) rotations in \(75.0 \mathrm{~s}\). How far does the pebble travel in that time?

A 2.00-m-long steel rod, pivoted at one end, swings in a vertical plane such that its lower end sweeps out an arc \(10.0 \mathrm{~cm}\) long. Determine the angle, in degrees and radians, through which the rod swings.

A box rests at a point \(2.0 \mathrm{~m}\) from the central vertical axis of a horizontal circular platform that is capable of revolving in the horizontal plane. The coefficient of static friction between box and platform is \(0.25 .\) As the rate of rotation of the platform is slowly increased from zero, at what angular speed will the box begin to slide?

A car has wheels each with a radius of \(30 \mathrm{~cm} .\) It starts from rest and (without slipping) accelerates uniformly to a speed of \(15 \mathrm{~m} / \mathrm{s}\) in a time of \(8.0 \mathrm{~s}\). Find the angular acceleration of its wheels and the number of rotations one wheel makes in this time. Remember that the center of the rolling wheel accelerates tangentially at the same rate as does a point on its circumference. We know that \(a_{T}=\left(v_{f}-v_{i}\right) / t\), and so $$a_{T}=\frac{15 \mathrm{~m} / \mathrm{s}}{8.0 \mathrm{~s}}=1.875 \mathrm{~m} / \mathrm{s}^{2}$$ Then \(a_{T}=r \alpha\) yields $$\alpha=\frac{a_{T}}{r}=\frac{1.875 \mathrm{~m} / \mathrm{s}^{2}}{0.30 \mathrm{~m}}=6.2 \mathrm{rad} / \mathrm{s}^{2}$$ Notice that we must introduce the proper angular measure, radians. Now use \(\theta=\omega_{i} t+\frac{1}{2} \alpha t^{2}\) to find $$\theta=0+\frac{1}{2}\left(6.2 \mathrm{rad} / \mathrm{s}^{2}\right)(8.0 \mathrm{~s})^{2}=200 \mathrm{rad}$$ and to get the corresponding number of turns divide by \(2 \pi\), $$ (200 \mathrm{rad})\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right)=32 \mathrm{rev} $$
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