Figure 11-35 shows two masses connected by a cord passing...

Key Concepts

Coupled System Dynamics

When components are connected, such as masses linked by a cord running over a pulley, the dynamics of each part become interdependent. The accelerations, forces, and torques acting on the different parts are linked by the constraints imposed by the connection. In such systems, the motion of one part (such as the sliding mass) directly impacts the motion of the other (like the hanging mass and the rotating pulley), requiring simultaneous application of equations from both rotational and translational dynamics to derive quantities such as acceleration and angular momentum.

Newton’s Laws for Rotational and Translational Dynamics

Newton’s laws are applied separately to the linear motion of the masses and the rotational motion of the pulley. For translational motion, Newton's second law (F = ma) is used to relate net forces to mass accelerations. In rotational motion, the analogous relation ? = I? (torque equals moment of inertia times angular acceleration) is used. When masses are connected by a cord that affects both translations of the masses and the rotation of the pulley, these laws must be combined to analyze the acceleration of the masses and the system’s angular momentum.

Angular Momentum

Angular momentum is a measure of the amount of rotation an object or system possesses, taking into account both the distribution of its mass relative to an axis and its angular velocity. For a point mass moving in a circular path, it is defined as the product of the mass, its linear speed, and the perpendicular distance to the axis. In a combined system where both point masses and an extended rigid body (like a pulley) are involved, the total angular momentum is the sum of the angular momentum of each component, and is conserved in the absence of external torques.

Moment of Inertia

The moment of inertia is a property of a rigid body that quantifies its resistance to changes in its rotational motion about an axis. It depends on the mass distribution of the body relative to the axis of rotation. In problems involving rotating pulleys, the moment of inertia is central to determining how the pulley’s rotation contributes to the system's overall dynamics, and it figures directly into equations such as those relating torque to angular acceleration.

Relationship Between Linear and Angular Quantities

The relationship between linear and angular kinematics is essential in analyzing systems where translation and rotation are coupled, such as with a cord moving over a pulley. This relationship is given by v = ?R, where v is the linear speed, ? is the angular speed, and R is the radius of the circular path. This connection allows one to convert between linear motion (of the masses) and rotational motion (of the pulley) when computing quantities like angular momentum and acceleration.

Transcript

00:01 So in this question we are given a situation as shown.

00:05 There is a table on which there's a mass m .a.

00:11 And a massless string attaches it to the around a pulley to another mass m .b, which hangs on the other side, from off the table.

00:21 And the pulley itself has a radius r0 and a moment of inertia of i.

00:29 And with this we are asked a few questions.

00:30 So the first part of question asks us to find the angular momentum of the entire system about the center point of the pulley or the axis of the axis of rotation of the pulley with the of the whole system the angular momentum which means that we will have to take angular momentum of this body moving at some velocity and this body moving at some velocity in this pulley rotating so we are told that we can find that as a function of the velocity of the blocks so let's say that there is a velocity v for this block and because there is no slip on the pulley and the string is massless and it is not extending it's not extending not and doesn't have any slack so we can say that the linear motion of the block a and the string is going to be translated to the second portion of the string and the block here such that it will also have the same velocity so with this we can say that the momentums of both these blocks become m a times v and mb times v respectively so since we have to find the moment of inertia sorry the angular momentum about the point about the axis of rotation of the pulley we take we have to take so remember the moment of inertia i'm sorry angular momentum of vector is r cross um momentum vector so in this case let's say the momentum vector here would be mv in this direction and the r vector perpendicular vector would be or the position vector let's say would be you can say it's this vector roughly r vector the position vector of the block center let's say to the mass so if you take a cross product of this you have to curl your fingers from the r vector to the momentum vector going from the smaller angle so in this case let's say it will be something like this this would be r vector and this would be mv vector.

02:46 So you have to curl your fingers in this direction, a smaller angle between r and mb vector.

02:50 And this would tell you that the angular momentum vector lies facing away from the plane of the paper or facing u.

03:00 So let's say that that is deposited direction.

03:03 So we'll just add the angular momentum for all these three.

03:06 So the first one would simply be m -a -v times r -0 because the perpendicular distance from mv about the point let's say this o point is r not the radius of the pulley because the momentum is going to be directed along the line of the string assuming the string is attached to the midline of the middle of the block so this will be the magnitude of the angular momentum of the first block and this will be the vector is pointing out of the page so we'll take that as positive similarly for the second one there'll be m bv vector will be m a and there'll be another r vector here such that you will have something like this you'll have r vector and you'll have a mbv vector such that if you take the r vector over here and curl your fingers from the smaller direction again you will see that again you will be curling your fingers in a way that your thumb points out of the page of the paper and here again, if you take the magnitude, it will still be mbv times r0, because again, the ma vector is r0 away from the center of the axis of the rotation of the pulley.

04:37 So these two values give you the same sign and the magnitude for the two blocks.

04:43 And for the for the pulley itself, we can say that the angular momentum of any rotating body is i -alpha.

04:54 Sorry i omega and so we can we have been given i and we can say it's i omega but we can write omega as v by r r not because there is again no slippage between the rope the rope and the the the pulley outer edge which means that the tangential velocity of the pulley is same as the velocity of the rope and so the omega has to be v by r0 at the edge of the pulley.

05:26 So we can actually remove omelmary v by r0.

05:30 So this becomes your answer you can simplify this.

05:32 This becomes m a r0 plus m br0 plus i by r0 times v so this is your answer for part a.

05:48 Now for part b let's assume that there is a let me remove some diagrams from here some vectors from here.

05:57 It's too much crowded here.

06:01 All right.

06:02 So let's say that now that this is the acceleration of the blocks.

06:09 This is the direction of acceleration.

06:10 We have to now find the acceleration of the two blocks.

06:13 Let's assume this is the direction and they also will be equal since velocities are equal for both...