QUESTION 10You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 76.4 kg, and its side-to-side width is 1.54 m. The knob is located 0.10 m from the right-hand side of the door. If you push with a force of 178 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) A1.70 rad/s^2 B6.78 rad/s^2 C1.19 rad/s^2 D4.24 rad/s^2 Hinge

Answered: Image /qna-images/answer/a10ebe67-d852-4aea-a589-3fb01dfe77f4.jpg

Answered: You push on a rectangular door at the…

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

QUESTION 10

You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 76.4 kg, and its side-to-side width is 1.54 m. The knob is located 0.10 m from the right-hand side of the door. If you push with a force of 178 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.)

A
1.70 rad/s^2

B
6.78 rad/s^2

C
1.19 rad/s^2

D
4.24 rad/s^2

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