The assembly of Prob. 18.67 is initially at rest (ω=0) when...

The assembly of Prob. 18.67 is initially at rest $(\omega=0)$ when a couple $\mathbf{M}_{0}$ is applied to axle $A B$. Knowing that the resulting angular acceleration of the assembly is $\boldsymbol{\alpha}=\left(150 \mathrm{rad} / \mathrm{s}^{2}\right) \mathbf{i}$, determine $(a)$ the couple $\mathbf{M}_{0},(b)$ the dynamic reactions at $A$ and $B$ immediately after the couple is applied.

Key Concepts

Dynamic Reaction Forces

Dynamic reaction forces are the forces exerted by supports or pivots due to the acceleration of a body. In systems undergoing rotational acceleration, these reactions are influenced by both the applied torques and the inertial effects caused by the body's mass distribution. Correctly determining these forces is crucial for ensuring the structural integrity of the supports in dynamic conditions.

Newton’s Second Law for Rotation

Newton’s second law for rotation states that the sum of torques acting on a rigid body is equal to the product of its moment of inertia and its angular acceleration. This principle is essential for analyzing systems involving rotational motion, as it links applied torques to the dynamic response of the body.

Torque and Couple

A torque (or couple) is a measure of the rotational force applied to a body. It results from two equal and opposite forces whose lines of action do not coincide, creating a turning effect. In rotational dynamics, the applied torque initiates or changes the rate of rotation of an object, directly influencing its angular acceleration.

Angular Acceleration

Angular acceleration is the rate at which an object's angular velocity changes over time. It is a key parameter in rotational motion, analogous to linear acceleration in translational motion, and is directly proportional to the net applied torque when divided by the moment of inertia.

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics that quantifies how mass is distributed relative to an axis of rotation. It determines how much torque is needed for a desired angular acceleration. A higher moment of inertia means the object is more resistant to changes in its rotational speed.

Transcript

00:01 Hello friends here in the problem it is given you should follow the diagram from the book assembly source consists of pieces of seat of aluminium of uniform thickness and total weight 7 .2 pounds and belted to the axle supported by the wearings a and b assembly is initially at rest when a couple m not is applied on the axel a .b knowing that angular acceleration is 150 radiant per second square i cap calculate couple apply on the accent and be dynamic reaction at a and b dynamic reactions mass of each sheet metal total mass is given so you may find mass of seat metal aluminum aluminum it is given 2 .7 upon circuit to point 2 .2 that is 0 .08 385 square upon fifth or you may write 0 .08385 slugs now sheet metal having the dimension b is equal to 6 inch that is 0 .5 fifth so area of assembly you can find area will be half v square see the figure v square there are two rectangular and two triangular right angle triangle it will be 3 v square 3 into 0 .5 square so you will get 0 .75 fit square so you can find the density so density of the seed you will get mass upon area mass per unit area you are finding mass is 0 .08385 divided by 0 .75 slugs per fit square and its value will be 0 .1 .1 .8.

04:07 0 .1 .8 slugs per fit square.

04:14 This is mass per unit area.

04:17 Remember it.

04:17 This is not mass per unit value.

04:19 It is mass per unit volume.

04:19 It is mass.

04:23 Per unit area remember it now moment of moments and product of inertia we will find moment and product of energy i mass can be written as row into area let us see here moment of energy of this portion separately i am drawing here geometrically let us see here just a moment let me this portion this is b this is a as you can say this is b this is b this is your x -axis this is your y -axis so i x this is b you can write each dimension is given b so ix you may write 1 by 3 b to the power 4 plus 1 by 3 so it becomes 2 by 3 of b bt power 4 so i x to v row into 1 by 2 by 3 bc power 4 substitute the value row already we have calculated density mass per unit area and v is 0 .5 to the power 4 so on solving it this would be 4 .658 10 to the power of minus 3 slug 6 similarly you can find i xy it can be written as b square into 3y 2b square b into half of v plus b square 5 by 2 b minus half b then you will get minus half v to the power 4 will be this is area now in terms of mass minus 1 by 4 row b key power 4 so you will get sorry this is 1 by 4 let me correct it first substitute the value density is 0 0 0 0718 0 .5 to the power 4 so on solving it this value will be minus 3 .4938 selects fit square.

08:48 Now this portion we have to find, i'm drawing here roughly...

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