/*! 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 7 What is the centripetal accelera... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the centripetal acceleration of a mass that moves in a circle of radius \(2.45 \mathrm{m}\) making 3.5 revolutions per second?

Short Answer

Expert verified
The centripetal acceleration is approximately 1183.33 m/s².

Step by step solution

01

- Find the angular velocity

The mass completes 3.5 revolutions per second. To find the angular velocity, use the formula \ \ \(\omega = 2\pi \times \text{{revolutions per second}} = 2\pi \times 3.5 = 7\pi \)\( \frac{\text{{rad}}}{{\text{{s}}}} \)
02

- Use the radius in the centripetal acceleration formula

The formula for centripetal acceleration is \ \ \(a_c = \omega^2 \cdot r\). \ Using \(r = 2.45 \mathrm{m}\) and \(\omega = 7\pi \frac{\text{{rad}}}{{\text{{s}}}} \), we have \ \ \(a_c = (7\pi)^2 \cdot 2.45\).
03

- Calculate the centripetal acceleration

Calculate \ \( (7\pi)^2 = 49\pi^2\). Then \ \(a_c = 49\pi^2 \times 2.45\). \ Approximating \(\pi \approx 3.14\), we get \( \pi^2 \approx 9.86 \). Therefore, \ \(a_c \approx 49 \times 9.86 \times 2.45 \approx 1183.33 \text{{ m/s}}^2\).

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Key Concepts

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

Angular Velocity
Angular velocity (\(\boldsymbol{u}\)) describes how fast an object rotates or revolves relative to another point. In other words, it tells us how quickly the angle of the object changes as it moves along a circular path. Mathematically, it is expressed in radians per second (rad/s).
To calculate angular velocity, use the formula: \(\boldsymbol{u = 2\boldsymbol{u} \times \text{{revolutions per second}} \text{{(in radians per second)}}}\).

In the given example, the mass completes 3.5 revolutions per second. Plugging in the values, we get \(\boldsymbol{u = 2\boldsymbol{u} \times 3.5 = 7\boldsymbol{u}} \text{{rad/s}}\). This tells us that the mass has an angular velocity of 7Ï€ radians per second as it moves along its circular path.
Circular Motion
Circular motion occurs when an object moves along the circumference of a circle. It can be uniform or non-uniform, depending on whether the object's speed remains constant or changes.

In this problem, the mass undergoes uniform circular motion, completing 3.5 revolutions per second at a constant rate. The key parameter of uniform circular motion is that the object's speed and the angular velocity are constant, which simplifies the calculations.

The radius of the circular path (2.45 meters in this case) is crucial in many calculations involving circular motion, such as finding the centripetal acceleration. 
If you know the angular velocity and the radius, you can easily compute the centripetal acceleration!
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, acting towards the center of the circle. Without it, the object would move in a straight line due to its inertia.

The formula to calculate the centripetal force (\boldsymbol{F_c}) is:
\[F_c = m \times a_c\]
where \boldsymbol{a_c} is the centripetal acceleration and \bold::m:: is the mass of the object. Centripetal force arises from various types of forces:
  • Tension in a string
  • Gravitational force
  • Electromagnetic force
In the solution provided, we've calculated the centripetal acceleration using the angular velocity (\boldsymbol{u}) and the radius (\boldsymbol{r}):\boldsymbol{a_c = u^2 \times r}By substituting \boldsymbol{u = 7u} rad/s and \bold::r:: = 2.45 m, we've derived the centripetal acceleration. Here, it helps to maintain the mass on its circular path.

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Most popular questions from this chapter

A satellite is in a circular orbit around the earth. The satellite turns on its engines so that a small force is exerted on the satellite in the direction of the velocity. The engines are on for a very short time and the satellite now finds itself in a new circular orbit. (a) State and explain whether the new orbit is closer to or further away from the earth. (b) Hence explain why the speed of the satellite will decrease. (c) It appears that a force, acting in the direction of the velocity, has actually reduced the speed. How do you explain this observation?

A rocket accelerates vertically upwards from rest with a constant acceleration of \(4.00 \mathrm{m} \mathrm{s}^{-2} .\) The fuel lasts for \(5.00 \mathrm{s}\) (a) What is the maximum height achieved by this rocket? (b) When does the rocket reach the ground again? (c) Sketch a graph to show the variation of the velocity of the rocket with time from the time of launch to the time it falls to the ground. (Take the acceleration due to gravity to be \(10.0 \mathrm{m} \mathrm{s}^{-2} .\)

Make velocity-time sketches (no numbers are necessary on the axes) for the following motions. (a) A ball is dropped from a certain height and bounces off a hard floor. The speed just before each impact-with the floor is the same as the speed just after impact. Assume that the time of contact with the floor is negligibly small. (b) A cart slides with negligible friction along a horizontal air track. When the cart hits the ends of the air track it reverses direction with the same speed it had right before impact. Assume the time of contact of the cart and the ends of the air track is negligibly small. (c) A person jumps from a hovering helicopter. After a few seconds she opens a parachute. Eventually she will reach a terminal speed and will then land.

An elevator starts on the ground floor and stops on the 10 th floor of a high- rise building. The elevator picks up a constant speed by the time it reaches the 1 st floor and decelerates to rest between the 9 th and 10th floors. Describe the energy transformations taking place between the 1 st and 9th floors.

A body of mass \(1.00 \mathrm{kg}\) is tied to a string and rotates on a horizontal, frictionless table. (a) If the length of the string is \(40.0 \mathrm{cm}\) and the speed of revolution is \(2.0 \mathrm{m} \mathrm{s}^{-1},\) find the tension in the string. (b) If the string breaks when the tension exceeds \(20.0 \mathrm{N},\) what is the largest speed the mass can rotate at? (c) If the breaking tension of the string is 20.0 N but you want the mass to rotate at \(4.00 \mathrm{m} \mathrm{s}^{-1},\) what is the shortest length string that can be used?

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