Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a rad

Gear A has a mass of 1 kg and a radius of gyration of 30 mm;...

Gear $A$ has a mass of $1 \mathrm{kg}$ and a radius of gyration of $30 \mathrm{mm}$; gear $B$ has a mass of $4 \mathrm{kg}$ and a radius of gyration of $75 \mathrm{mm}$; gear $C$ has $\mathrm{a}$ mass of $9 \mathrm{kg}$ and a radius of gration of $100 \mathrm{mm}$. The system is at rest when a couple $\mathrm{M}_{0}$ of constant magnitude $4 \mathrm{N} \cdot \mathrm{m}$ is applied to gear $C .$ Assuming that no slipping occurs between the gears, determine the number of revolutions required for disk $A$ to reach an angular velocity of $300 \mathrm{rpm} .$

Key Concepts

Moment of Inertia

In rotational dynamics, the moment of inertia quantifies an object's resistance to angular acceleration and is analogous to mass in linear motion. It depends on both the object's mass and the distribution of that mass relative to the axis of rotation. The radius of gyration is used to simplify this concept by providing an equivalent distance at which the total mass can be assumed to be concentrated, making it straightforward to compute the moment of inertia as the product of mass and the square of the radius of gyration.

Torque and Angular Acceleration

Torque, or the applied couple, is the rotational equivalent of force and causes angular acceleration in a rotating system. According to the rotational form of Newton's second law, the angular acceleration of an object is determined by the net applied torque divided by its moment of inertia. This fundamental relationship allows one to analyze how a constant applied torque will increase the angular velocity over time.

Angular Kinematics

Angular kinematics involves the study of rotational motion without using the forces that cause the motion. It includes relationships between angular displacement, angular velocity, and angular acceleration. When an object experiences constant angular acceleration, kinematic equations can be used to relate the final angular velocity to the number of revolutions made, which is essential for determining the response of a rotating system to a constant applied torque.

Gear Ratio and No-slip Condition

When multiple gears are in contact and no slipping occurs between them, their angular displacements, velocities, and accelerations are geometrically related by the gear ratio. The condition of no slip ensures that the point of contact between the gears moves with the same linear speed, and this relationship allows the dynamics of one gear to be translated into the dynamics of interconnected gears. This concept is crucial in analyzing how an applied torque on one gear affects the rotation of another gear in the system.

Transcript

00:02 Hi everyone, as shown in the figure, gear a having the mass 1 kg and radius of gyration 30mm that is 0 .03 meter.

00:25 Mass of gear v is given 4kg and radius of duration 75mm that is 0 .75 meters.

00:47 Mass of gear c is given 9 kg and its radius of variation is 100 mm that is 0 .1 meter.

01:04 Radius of each gear are as shown in the figure the system is initially at rest so initial velocity or angular velocity is 0.

01:21 Moment of couple am not having the magnitude 4 newton into meter.

01:27 Applied on c there is no silping occurs between the gap that is there is a pure rolling motion we have to calculate number of revelation required for this a to reach to reach angular velocity of 300 r pf let us start solving it first we will calculate moment of inertia of each gap this problem is based on work energy so first we are finding moment of inertia of each gap.

02:46 So calculation of moment of inertia first for gap a.

02:57 Formula to be used product of mass into square of radius of variation.

03:04 General formula is m into k square.

03:07 We are using it for gap a.

03:09 Mass of a is given 1 kg and its radius of variation is 0 .03.

03:16 So you will get 0 .9 into 10 to the power of minus 3 kg meter square and that of v would be mass of v is given 4 kg and its radius of variation is 0 .075.

03:45 So moment of inertia of v you will get 22 .5 into 10 to the power minus 3 kg meter square.

03:57 That of c you will get 9 mass of c is 9 kg and its radius of variation is 0 .1 so you will get 90 into 10 to the power of minus 3.

04:21 Now we will study the kinematics of it analysis of kinematics are a radius of the gear a is 50mm meters radius of in the variation of the inner and outer two, there are two girls.

04:58 For gear we have r1 to be 100mm that is outer one and r2, and r2, 50mm.

05:22 This is for a and for gerc radius is rc and it is given 150mm at a of content between there a and v, we can write r1 omega v, ra, omega, a, so omega -v, you will get ra upon r1, substituting the value, ra is given 50mm, r1, r1 is 100 millimeter into angular velocity of a so we will find angular velocity of v is half of angular velocity of a or you may write 0 .5 times a so this is the one of the relation we have obtained similarly at content point between gear b and c we can write rc omega c to v r2 omega b that is angular velocity of c will be r2 upon rc into omega v substitute in the value.

07:58 R2 is 0 .5, sorry, 50 mrc is 150.

08:10 Omega v is 0 .5 into omega a.

08:15 So omega c you will get 0 .1667 omega a...

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