Answered: If the torque required to remove the…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter10: Rotational Motion And Angular Momentum

Section: Chapter Questions

Problem 15PE: Consider the 12.0 kg motorcycle wheel shown in Figure 10.38. Assume it to be approximately an...

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ques: 1) If the torque required to remove the nut is 33.8 N-m, what force (in N) must be exerted on the end of the wrench to loosen the nut? Assume the wrench has a length of 28.2 cm and the force is applied perpendicularly.

ques2) A wheel rotates with a constant angular acceleration. It has an angular velocity of 3.2 rad/s at time t = 0 s, and 2.0 s later has an angular velocity of 19.2 rad/s. What is the angular displacement (in rad) of the wheel during the time interval?

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 .

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Step 1 :Introduction

Since you have posted multiple questions we will solve the first question for you. If you want any specific question to be solved please specify the question number or post that question alone.

The torque acting on an object is calculated by using the formula $τ = F d \sin (θ)$ , where F is the force acting , $d = 0.282 m$ is the distance from the pivot point and $θ = 90 °$ as the force is acting perpendicularly.

Step 2 :Calculation of the force

Calculate the required force by using the formula $τ = F d \sin (θ)$ , where $τ = 33.8 N m$ is the torque acting , $d = 0.282 m$ is the distance from the pivot point and $θ = 90 °$ as the force is acting perpendicularly.

$\begin{array}{rcl} τ & = & F d \sin (θ) \\ F & = & \frac{τ}{d \sin (θ)} \\ = & \frac{33.8 N m}{(0.282 m) \times \sin (90 °)} \\ \approx & 120 N \end{array}$

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