Each of the gears A and B has a mass of 2.4 kg and a radius of gyration of 60 mm, while gear C h

step 1 Okay, so it's probably have three gears. So first of all, let's try to draw the gears in here. S

Each of the gears A and B has a mass of 2.4 kg and a radius...

Key Concepts

Transmission of Torque and Tangential Forces in Gears

In gear systems, the applied torque is transmitted between gears, and the resulting tangential force at the point of contact is related to the torque and the radius of the gear by the relationship F = Torque/Radius. This concept is important for understanding how forces are converted and transmitted in mechanical systems involving rotating elements.

Angular Kinematics

Angular kinematics refers to the equations that relate angular displacement, angular velocity, angular acceleration, and time in rotational motion. These equations are analogous to linear kinematics and are used to calculate quantities such as the number of revolutions required to change the angular velocity over a specified interval.

Moment of Inertia

The moment of inertia quantifies an object's resistance to changes in its rotational motion. It is determined by both the mass of the object and its distribution relative to the axis of rotation, typically calculated using the relation I = m·k², where k is the radius of gyration. This concept is crucial when analyzing how an applied torque (or couple) affects the angular acceleration of rotating bodies.

Newton’s Second Law for Rotation

Newton’s second law for rotation establishes the relationship between the net torque acting on a body, its moment of inertia, and the resulting angular acceleration, given by the equation M = I·?. This fundamental principle allows one to determine how an applied couple will change the rotational speed of an object.

Transcript

00:01 Okay, so it's probably have three gears.

00:04 So first of all, let's try to draw the gears in here.

00:09 So you have the two gears in here and the third one in here.

00:14 This one is a little bit bigger than the other two.

00:23 So that's basically our system.

00:27 Okay, what we know.

00:29 This is going to be the gear a, b, and c.

00:35 We know the problem says to us that the mass, let's put here, the mass of gear a is going to be equal the mass of gear b, which is equal 2 .4 kilograms.

00:58 What else do we know? we know that the radius is 60 millimeters.

01:05 So we have the radius a equals the radius b that is equal 60 millimeters.

01:18 What else do we know? we know that the mass of gear c is 12 kilograms.

01:31 And we also know that the radius, the radius of duration of the gear c, is 150 millimeters okay so that's basically the information about the system that we have we also know we also know that a couple m so let's put here we have a couple m with constant magnitude of 10 newtons meters is applied to the gear c and this problem just want to know in the first part so in part a we want to know the number of revolutions on the gear c require for its angular speed increases from let's put here the angular speed this is omega -1 goes to omega -2 and we want that omega 1 is equal 100 to let's see 450 rotations per minute so that's part a we want to know the number or revolutions so we want theta the number of revolutions for the speed increase by this amount and in part b we want the corresponding tangential force acting on the gear a.

03:20 So we want the tangential force.

03:28 Okay, first of all, we know that the three gears has a moment of inertia.

03:36 So let's put here the moment of inertia of the gear a, which is going to be called the gear b because they are equal.

03:46 And we know that the moment of inertia is just m r square which gives us 2 .4 they multiplies in meters 0 .06 square okay that gives us a total of 8 .64 times 10 to the minus 3 kilograms meters square so that's gives us a total of 8 .64 times 10 to the minus 3 kilograms meters square so that's the moment over inertia of the gear a and the gear b.

04:26 Now let's do the same for the gear c.

04:30 The moment of inertia of the gear c is the same equation so it's just 12 that multiplies the radius which is 0 .15 square.

04:44 That gives us a total of 270 times 10 to the minus 3 kilograms meters square.

04:57 Okay that's the first thing we must know the moment of inertia of each one of the gears but now in order to calculate the part a we want to know the number of revolutions to the speed increase and to calculate this we must use the conservation of the energy let's put here on black we must use the conservation of energy therefore we know that the connecting energy of the system in the first position plus the energy generated by the couple to go from position one to position two needs to be equal the connecting energy and position two okay therefore we must calculate each one of these quantities and after that we will find the answer for part a which is the number of rotations so first of all who is the connecting energy in the first situation is just the connecting energy of the gear a in the first position plus the connecting energy of the gear b in the first position plus the connecting energy of the gear c in the first position and that's just let me see half of i that multiplies omega let's put in this way this is just half and multiplies i a omega -a plus i b omega b plus i c omega c okay but how we're going to we already know what is the values of the moment of inertia we just calculate this here but what is the values of the angular speed okay we know the angular speed of the gear c first one needs to be 100 okay therefore let's use the kinematics and we know that r, let's put here in other color, we know that ra, omega, a, needs to be equal rc, omega, c.

07:46 Therefore, we can say that omega -a is just rc, omega -c, divided by r -a.

08:00 And that gives us an angular speed, of let's me see 150 divided by 60 omega c which gives us a total of 2 .5 omega c and that's the correlation between omega a and omega c therefore in the first position what we're going to call omega a 1 is going to be 2 .5 times 100 which gives us a total of, let me see, 25, 250, actually, 250.

08:54 Okay, rotations per minute.

08:59 Okay, so that's the value of omega a1, which is going to be equal the omega b1.

09:05 And now we can calculate the connecting energy in the first position.

09:12 Let's just make one observation here that we do not want to calculate our energy in rotations per minute.

09:21 We want in radiance per second.

09:24 Therefore this is going to be simplified.

09:27 Let's put here arrow.

09:30 This is going to be 25 divided by 3 pi rads per second.

09:41 Okay and that's the speed that we're going to use so we just need to substitute the numerical values in here and we're going to find that k1 is just after using these values that we have for the moment of inertia a b and c and the value for this speed we will find that the connecting energy in the first position is going to be equal 20 .7 to 6 joules.

10:27 So that's the energy of the first position.

10:32 In the second position, well, in the second position, let's make a division here, we again needs to use the same equation, but this time we substitute the ones for twos.

10:47 And again we're going to see the same problem with the angular speed.

10:54 We need to find the angular speed of the gear a and the gear b.

10:59 And we're going to use the same calculation.

11:02 So let's do it here.

11:05 We know that the correlation between omega a and omega c is given by this equation here.

11:15 2 .5 omega -c and in the second position we will have that omega -a -2 is going to be 2 .5 that multiplies 450 rotations per minute.

11:33 This gives us precisely, let me see 37...

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