Suppose a yo-yo has a center shaft that has a 0.21 cm radius and that its string is being pulled.Variables: r1 = 0.21 cmr2 = 3.9 cma = 1.9 m/s2t = 0.75 spart a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.9 m/s2, what is the angular acceleration of the yo-yo in rad/s2? part b) What is the angular velocity after 0.75 s if it starts from rest in rad/s? part c) The outside radius of the yo-yo is 3.9 cm. What is the tangential acceleration of a point on its edge in m/s2?

Answered: Image /qna-images/answer/cbced347-3c74-4a03-8d9d-75fe236c90b5.jpg

Answered: Suppose a yo-yo has a center shaft that…

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

Suppose a yo-yo has a center shaft that has a 0.21 cm radius and that its string is being pulled.

Variables:

r₁ = 0.21 cm
r₂ = 3.9 cm
a = 1.9 m/s²
t = 0.75 s

part a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.9 m/s², what is the angular acceleration of the yo-yo in rad/s²?

part b) What is the angular velocity after 0.75 s if it starts from rest in rad/s?

part c) The outside radius of the yo-yo is 3.9 cm. What is the tangential acceleration of a point on its edge in m/s²?

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 with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

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