The 1.5-kg uniform slender bar AB is connected to the 3-kg...

Name: Problem 35, Chapter 17: Vector Mechanics for Engineers: Statics and Dynamics
Uploaded: 2020-09-25T02:48:07-08:00
Duration: 10 min 49 s
Channel: Khoobchandra Agrawal
Description: The 1.5-kg uniform slender bar AB is connected to the 3-kg gear B that meshes with the stationary outer gear C. The centroidal radius of gyration of gear B is 30 mm. Knowing that the system is released from rest in the position shown, determine (a) the angular velocity of the bar as it passes through the vertical position, (b) the corresponding angular velocity of gear B.

The 1.5-kg uniform slender bar AB is connected to the 3-kg gear B that meshes with the stationary outer gear C. The centroidal radius of gyration of gear B is 30 mm. Knowing that the system is released from rest in the position shown, determine (a) the angular velocity of the bar as it passes through the vertical position, (b) the corresponding angular velocity of gear B.

Key Concepts

Conservation of Mechanical Energy

This principle states that in the absence of non-conservative forces (such as friction), the total mechanical energy of a system (the sum of potential and kinetic energies) remains constant. In the context of rotational systems, gravitational potential energy can be converted into rotational kinetic energy as the system moves from one position to another.

Rotational Kinetic Energy

Rotational kinetic energy is the kinetic energy due to the rotation of an object and is given by the expression (1/2)I?², where I is the moment of inertia and ? is the angular velocity. This concept is important because it quantifies the energy involved in rotating bodies and appears in energy conservation calculations.

Moment of Inertia and Radius of Gyration

The moment of inertia is a measure of an object's resistance to changes in its rotational motion, dependent on both its mass and how that mass is distributed relative to the axis of rotation. The radius of gyration is a related concept that represents the distance from the axis at which the entire mass of a body can be assumed to be concentrated to yield the same moment of inertia.

Gear Mesh and Angular Velocity Relationship

In gear systems, the condition of meshing enforces a geometric relationship between the angular displacements and velocities of interacting gears. This relationship is determined by the ratio of the radii or number of teeth of the gears, ensuring that the tangential velocities at the point of contact are equal.

Slender Bar Dynamics

A slender bar is treated as a rigid body with a uniform mass distribution, which simplifies the calculation of its moment of inertia about an axis. This concept is useful in problems involving rotational dynamics of beams and bars, where approximations based on uniform mass distribution are valid.

Transcript

00:01 There is a uniform slender bar a .v having the mass 1 .5 kg and length 120mm connected to the gear b of mass 3 kg and off radius 50mm.

00:26 The central radial radius of the gear b is 30mm.

00:40 The system is initially at rest.

00:43 So omega 1 is 0.

00:45 We have to calculate angular velocity of the bar as it is 30.

00:49 Passes to the vertical position and corresponding angular velocity of gear v that we have to calculate.

00:58 Let us first discuss the kinematics of it.

01:15 Velocity of v can be defined as length of av into angular velocity of av.

01:38 So angular velocity of gear v will be l -a -v -o -a -v -o -mega -a -v upon its radius that is r b velocity of rod a b will be half of velocity of the moment of inertia we will find moment of inertia of the rod a v mass of a square of its length 112 and that of gear in the shape of the disc b will be mass of disc into its square of radiation of variation kinetic energy of the system, t is equal to kinetic energy of the rod a b.

03:29 So it having translatory kinetic energy as well as rotational kinetic energy plus kinetic energy of the disk be half mb.

04:01 Now substituting the value in terms of omega a b of each, then we will get half of 1 by 4 mark.

04:19 Of av 1 by 12 of mass of av mass of v plus k square upon r v into l square a v into angular velocity of a v square now substituting the value mass of rod a v is given 1 .5 kg mass of disc v is given 3 kg radius of duration of disk v is given 30mm and radius is given 50mm length of ab is given 0 .12 meter so kinetic energy you will get 0 .0 3297 6 omega square a v for position 1 as shown in the figure gravitation potential energy is zero and kinetic energy is also zero.

06:41 For position two, point b will come out just directly below a.

07:03 So gravitational potential energy will be weight of av into lav by 2 minus weight of v to lav substitute the value.

07:32 Mass of rod av into g into g, that is the weight, length of av is given 0 .12 divided by 2 plus weight of the gap v...

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