Two uniform disks of the same material are attached to a shaft as shown. Disk A has a weight of

Two uniform disks of the same material are attached to a...

Key Concepts

Angular_Kinematics

Angular kinematics involves the set of equations that relate angular displacement, angular velocity, angular acceleration, and time. These relationships, analogous to linear kinematics, are essential for predicting the behavior of rotating systems under constant angular acceleration, such as determining the number of revolutions required to reach a specified angular velocity from rest.

Moment_of_Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. For a uniform disk, the moment of inertia depends on its mass and the square of its radius, typically expressed by the formula I = 1/2 m r². In problems where the geometry or mass distribution changes (for example, by altering the thickness of a disk), it directly influences how the disk responds to applied torques.

Torque

Torque is the rotational equivalent of force and quantifies the tendency of a force to produce angular acceleration around an axis. It is calculated as the product of the force and the distance from the axis at which it is applied. In dynamics problems, the net applied torque determines the angular acceleration of the system via Newton’s second law for rotation.

Work_Energy_Principle_in_Rotation

In rotational dynamics, the work done by a torque over a certain angular displacement is converted into rotational kinetic energy. This work-energy relationship allows one to set the work done equal to the change in rotational kinetic energy, thus linking the input torque and the resultant change in angular velocity. It is especially useful when analyzing systems where the torque is applied over a specific angular distance.

Transcript

00:01 Hello friends here it is given that two uniform disk of same material means density of both are same this is the disc a this is the disc b so density of both is same are attached to the soft as shown in the figure the disc a having the weight weight of the disc is given 10 pound and its radius is 6 inch radius of b is given n into r a that is n into six inch thickness of b is twice of b if moment of couple is applied on the disc a having the magnitude 22 pound fit starting from rest then in five revelation its velocity angular velocity becomes 480 then you have to find the radius of disk v.

01:41 Let us start solving it.

01:43 So this is given.

01:44 This problem is based on work energy theorem.

01:50 Work done by moment on the disk is equal to change in its kinetic energy.

01:55 So let us start solving it.

01:58 First we have to calculate moment of inertia of both the disk.

02:03 So mass of disk a is will be weight.

02:08 Divided by acceleration due to gravity it is given 10 pounds divided by 32 .2 fifth per second square so mass of a we will get 10 upon 32 .2 slugs now moment of inertia of disk a m a into r a square divided by 2 mass is 10 upon 32 .2 radius is 6 inch be required in fit so this value you will get 03882 select fit square now moment of inertia of v so first we need to calculate the mass mass of b will be by n r a square thickness that is 2b into row and mass of a will be m r a square thickness into row dividing both you will get 2 into n square so mass of v disk is 2 n squared times of mass of a now we will obtain its moment of inertia so moment of inertia of b will be half mb into r v square square mass of a is 10 pound divided by 32 .2.

04:42 Its radius is 6 into n squared.

04:47 So this would be equal to 07764 n to the power 4 select 6 squared.

05:14 This is moment of inertia of disk v.

05:22 So total moment of energy of the system will be ia plus iv combined moment of inertia of the both that is 0382 plus 0 .0764 n to the power 4...

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