Answer Question 5:5. The merry-go-round above slows down and comes to a stop in Δt = 6.00 s. What is the average angular acceleration during this time? 2. A merry-go-round goes around one complete rotation in time T= 9.00 s. What is the angular velocity, o, andthe frequency, f, of the merry go round?Time taken to complete one revolution ; T=9 secAs we know°, time required to complete one revolutioncalled as time period.Hence time period T=9 secangular frequency is given by relation;W= 2入W = 0.698 rad/s Answerfrequency (f) given as0.698こニ2てf = 0.111 HzAnswer 3. Calculate the linear velocity, v, of horse on that merry go round that is located a distance r= 2.50 m from thecenter.Vニrdr- distunce from centerW-Angular Speed.Y= 2.50mW= 0.698 rad/sV= (2.50m) ×V= 1.745 m/sLinear velocity, v= 1.745 m/s(0.698 rad/s)

Initial angular velocity= "wi= 0.698 rad/ s final angular velocity ="wf"= 0rad/s (as stops) Time…

Answered: The merry-go-round above slows down and…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Concept explainers

Question

Answer Question 5:

5. The merry-go-round above slows down and comes to a stop in Δt = 6.00 s. What is the average angular acceleration during this time?

2. A merry-go-round goes around one complete rotation in time T= 9.00 s. What is the angular velocity, o, and
the frequency, f, of the merry go round?
Time taken to complete one revolution ; T=9 sec
As we know°, time required to complete one revolution
called as time period.
Hence time period T=9 sec
angular frequency is given by relation;
W= 2入
W = 0.698 rad/s Answer
frequency (f) given as
0.698
こ
ニ
2て
f = 0.111 Hz
Answer

3. Calculate the linear velocity, v, of horse on that merry go round that is located a distance r= 2.50 m from the
center.
Vニrd
r- distunce from center
W-Angular Speed.
Y= 2.50m
W= 0.698 rad/s
V= (2.50m) ×
V= 1.745 m/s
Linear velocity, v= 1.745 m/s
(0.698 rad/s)

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Angular speed, acceleration and displacement

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions