How can i solve this exercise? 3. A rod is released from rest at an angle of 45° and is made to rotate about a pivotpoint as shown below. It has a length 30 cm and mass 300 g. What is its angularvelocity at its lowest point?L = 30 cmm = 300 g%3D

Answered: 3. A rod is released from rest at an…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter7: Rotational Motion And Gravitation

Section: Chapter Questions

Problem 1CQ: A disk rotates about an axis through its center. Point A is located on its rim and point B is...

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How can i solve this exercise?

3. A rod is released from rest at an angle of 45° and is made to rotate about a pivot
point as shown below. It has a length 30 cm and mass 300 g. What is its angular
velocity at its lowest point?
L = 30 cm
m = 300 g
%3D

Expert Solution

Step 1

The law of conservation of energy in rotational motion states that sum of rotational kinetic energy and potential energy remains constant.

${KE}_{rot} + PE = constant$

Rotational kinetic energy is given by,

${KE}_{Rot} = \frac{1}{2} {Iω}^{2}$

Where,

$\begin{array}{rcl} I & = & Moment of inertia \\ ω & = & Angular velocity \end{array}$

Potential energy is given by,

$PE = mgh$

Where,

$\begin{array}{rcl} m & = & Mass \\ h & = & Height from the reference \\ g & = & Acceleration due to gravity \end{array}$

Step 2

Where $CM$ repesents the center of mass of the rod.

The potential energy at the initial position is converting in to the rotational kinetic energy of the rod at lowest position. Applying law of conservation of energy,

$mgh = \frac{1}{2} {Iω}^{2}$ ...(1)

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