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Chapter 9: Problem 27

A mass of \(1.5 \mathrm{~kg}\) out in space moves in a circle of radius \(25 \mathrm{~cm}\) at a constant \(2.0 \mathrm{rev} / \mathrm{s}\). 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) \(39.478\) m/s虏, (c) \(59.217\) N.

Step by step solution

01

Convert Units

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

Calculate Tangential Speed

The tangential speed \( v \) is calculated using the formula:\[ v = 2\pi r f \]where \( f \) is the frequency in revolutions per second. Here, \( f = 2.0 \text{ rev/s} \) and the radius is \( r = 0.25 \text{ m} \). Substitute the values:\[ v = 2\pi \times 0.25 \times 2 = \pi \text{ m/s}\]
03

Calculate Centripetal Acceleration

The centripetal acceleration \( a_c \) is calculated using:\[ a_c = \frac{v^2}{r} \]Substitute the values of \( v = \pi \text{ m/s} \) and \( r = 0.25 \text{ m} \):\[ a_c = \frac{\pi^2}{0.25} \approx 39.478 \text{ m/s}^2 \]
04

Calculate Centripetal Force

The centripetal force \( F_c \) required to maintain the motion is given by:\[ F_c = m \times a_c \]where \( m = 1.5 \text{ kg} \) is the mass. Substitute the values:\[ F_c = 1.5 \times 39.478 \approx 59.217 \text{ N} \]
05

Conclusion

The results accurately depict the motion of the mass: (a) the tangential speed is \(\pi\) m/s, (b) the acceleration is approximately \(39.478\) m/s\(^2\), and (c) the required centripetal force is approximately \(59.217\) 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 is the speed at which an object moves along the path of a circle. It can be thought of as the linear speed along the circular path, and it changes depending on how fast the object is rotating and the radius of the circle. In our example, the object completes 2 revolutions per second. Since one complete revolution is a full circle of motion, the frequency of rotation helps determine the tangential speed. Tangential speed ( v ) is calculated by the formula: \[ v = 2\pi r f \]
  • \(r\) is the radius of the circle (in meters).
  • \(f\) is the frequency of rotation (in revolutions per second).
In this case, the object's radius is 0.25 meters and the frequency is 2 revolutions per second. By plugging these into the formula, we find that its tangential speed is approximately \(\pi\) m/s. This speed is the rate at which the object travels around the edge of the circle. Simplifying complex motion into straightforward variables like these helps comprehend circular dynamics clearly.
Centripetal Acceleration
Centripetal acceleration occurs when an object moves in a circular path, causing it to constantly change direction. It is an acceleration that points towards the center of the circle, keeping the object moving along the circular path instead of going straight. This is a crucial reason why cars can navigate curves safely or planets remain in orbit. The formula to calculate centripetal acceleration ( a_c ) is: \[ a_c = \frac{v^2}{r} \]
  • \(v\) is the tangential speed (in m/s).
  • \(r\) is the radius of the circle (in meters).
Using our earlier result where \(v = \pi\) m/s, and knowing the radius is 0.25 meters, we compute the centripetal acceleration to be around 39.478 m/s虏. This value tells us how quickly the object's speed is changing as it continually redirects its motion inward, maintaining the curve.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path. It is not a new kind of force but usually supplied by another force like gravity or tension. This force is essential to maintaining circular motion and is directed towards the circle's center. We can calculate the centripetal force ( F_c ) using the formula: \[ F_c = m \times a_c \]
  • \(m\) is the mass of the object (in kilograms).
  • \(a_c\) is the centripetal acceleration (in m/s虏).
For the object with a mass of 1.5 kg and previously calculated acceleration of about 39.478 m/s虏, the centripetal force required is approximately 59.217 Newtons. This quantifies the energy needed to redirect the object continuously in its circular path. Understanding this force helps in designing systems ranging from fairground rides to satellite orbits efficiently.

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Most popular questions from this chapter

( \(a\) ) Compute the radial acceleration of a point at the equator of the Earth. (b) Repeat for the North Pole of the Earth. Take the radius of the Earth to be \(6.37 \times 10^{6} \mathrm{~m}\).

Express each of the following in terms of other angular measures: (a) \(28^{\circ}\), (b) \(\frac{1}{4} \mathrm{rev} / \mathrm{s}\), (c) \(2.18 \mathrm{rad} / \mathrm{s}^{2}\). (a) \(28^{\circ}=(28\) deg \()\left(\frac{1 \mathrm{rev}}{360 \mathrm{deg}}\right)=0.078 \mathrm{rev}\) \(=(28 \mathrm{deg})\left(\frac{2 \pi \mathrm{rad}}{360 \mathrm{deg}}\right)=0.49 \mathrm{rad}\) (b) \(\begin{aligned} \frac{1}{4} \frac{\mathrm{rev}}{\mathrm{s}} &=\left(0.25 \frac{\mathrm{rev}}{\mathrm{s}}\right)\left(\frac{360 \mathrm{deg}}{1 \mathrm{rev}}\right)=90 \frac{\mathrm{deg}}{\mathrm{s}} \\ &=\left(0.25 \frac{\mathrm{rev}}{\mathrm{s}}\right)\left(\frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}}\right)=\frac{\pi}{2} \frac{\mathrm{rad}}{\mathrm{s}} \end{aligned}\) (c) \(\begin{aligned} 2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}} &=\left(2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\right)\left(\frac{360 \mathrm{deg}}{2 \pi \mathrm{rad}}\right)=125 \frac{\mathrm{deg}}{\mathrm{s}^{2}} \\ &=\left(2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\right)\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right)=0.347 \frac{\mathrm{rev}}{\mathrm{s}^{2}} \end{aligned}\)

A \(200-\mathrm{g}\) object is tied to the end of a cord and whirled in a horizontal circle of radius \(1.20 \mathrm{~m}\) at a constant \(3.0 \mathrm{rev} / \mathrm{s}\). Assume that the cord is horizontal - that is, that gravity can be neglected. Determine ( \(a\) ) the centripetal acceleration of the object and ( \(b\) ) the tension in the cord. (a) The object is not accelerating tangentially to the circle but is undergoing a radial, or centripetal, acceleration given by $$ a_{c}=\frac{v^{2}}{r}=r \omega^{2} $$ where \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). Since \(3.0 \mathrm{rev} / \mathrm{s}=6.0 \pi \mathrm{rad} / \mathrm{s}\), $$ a_{C}=(1.20 \mathrm{~m})(6.0 \pi \mathrm{rad} / \mathrm{s})^{2}=426 \mathrm{~m} / \mathrm{s}^{2}=0.43 \mathrm{~km} / \mathrm{s}^{2} $$ (b) To cause the acceleration found in \((a)\), the cord must pull on the \(0.200-\mathrm{kg}\) mass with a centripetal force given by $$ F_{C}=m a_{C}=(0.200 \mathrm{~kg})\left(426 \mathrm{~m} / \mathrm{s}^{2}\right)=85 \mathrm{~N} $$ This is the tension in the cord

Through how many radians does a point fixed on the Earth's surface (anywhere off the poles) move in \(6.00 \mathrm{~h}\) as a result of the Earth's rotation? What is the linear speed of a point on the equator? Take the radius of the Earth to be \(6370 \mathrm{~km}\).

A car moving at \(5.0 \mathrm{~m} / \mathrm{s}\) tries to round a corner in a circular arc of \(8.0 \mathrm{~m}\) radius. The roadway is flat. How large must the coefficient of friction be between wheels and roadway if the car is not to skid?
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