An Atwood’s machine consists of blocks of masses m1=10.0 kg and m2=20.0 kg attached by a cord an

step 1 T sub 2 must be greater than T sub 1. This is 4 part A. And this is true because first, they mus

An Atwood’s machine consists of blocks of masses m1=10.0 kg...

An Atwood’s machine consists of blocks of masses $m_{1}=10.0 \mathrm{kg}$ and $m_{2}=20.0 \mathrm{kg}$ attached by a cord and $m_{2}=20.0 \mathrm{kg}$ attached by a cord running over a pulley as in Figure P8.46. The pulley is a solid cylinder with mass $M=8.00 \mathrm{kg}$ and radius $r=0.200 \mathrm{m} .$ The block of mass $m_{2}$ is allowed to drop, and the cord turns the pulley without slipping. (a) Why must the tension $T_{2}$ be greater than the tension $T_{1} ?$ (b) What is the acceleration of the system, assuming the pulley axis is frictionless? (c) Find the tensions $T_{1}$ and $T_{2}$ .

Key Concepts

Moment of Inertia

The moment of inertia is a measure of an object’s resistance to changes in its rotational motion, depending on how its mass is distributed relative to the axis of rotation. In the context of a pulley with a significant mass, its moment of inertia plays a critical role in the system's dynamics by determining how much torque is needed to achieve a given angular acceleration, thereby affecting the tension differences in the connecting cord.

Constraint Relationships in Pulley Systems

These are relationships that connect the linear motion of the masses to the rotational motion of the pulley through the condition of no slipping. Specifically, the linear acceleration of the masses is directly related to the angular acceleration of the pulley (a = r?). This constraint is crucial for linking the translational and rotational equations of motion in the analysis of such systems.

Newton's Second Law (Translational Dynamics)

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In problems involving masses connected by a cord, it is applied individually to each mass to determine their linear accelerations and required forces (or tensions) acting in their respective directions.

Newton's Second Law (Rotational Dynamics)

This extension of Newton's second law applies to rotating bodies and relates the net torque acting on a body to its moment of inertia and angular acceleration (? = I?). In pulley systems, this law explains how a difference in tensions across the pulley creates a net torque that results in the angular acceleration of the pulley.

Transcript

00:01 T sub 2 must be greater than t sub 1.

00:05 This is 4 part a.

00:06 And this is true because first, they must be different in order to provide a net torque.

00:11 So we know that the net torque is not equaling 0.

00:14 And so that's why this needs to occur.

00:17 We know that the angular acceleration is also not going to equal 0, depending on what we choose positive and negative to be, either counterclockwise or clockwise.

00:31 And we know that the system will, if the system is rotating clockwise, so system rotating clockwise, this means that this is also necessary for t sub 2 to be greater than t sub 1.

00:57 And this is when m sub 2 is greater than m sub 1.

01:02 So essentially, because of that, t .72 has to be greater than t sub 1.

01:08 Also, there must be, again, there must be a net torque.

01:22 This would be into the page.

01:30 That would be your answer for part a.

01:32 And for part b, we are going to say that clockwise is positive.

01:43 So given this, we can then apply newton's second law for each option.

01:47 So we can say that first for m sub 1 we can say the sum of forces in the y direction equals m sub 1 times the acceleration and this is going to be equalling to t sub 1 minus m sub 1 g and so we can then say that t sub 1 is equaling m sub 1 multiplied by g plus a the acceleration.

02:18 We can say for m sub 2 a continuing 1 for part b m sub 2 we can say that then the sum of the forces in the y direction is equalling m sub 2a this would be equalling m sub 2 g minus t sub 2 giving us t sub 2 equaling m sub 2 multiplied by g minus a so as you can see very similar however we have g minus a g plus a at this point we know that for m the pulley the mass of the pulley itself we can say that the sum of the torques is going to equal the moment of inertia times the angular acceleration.

02:59 And this is going to be equalling r times t sub 2 minus r times t sub 1, equalling the moment of inertia times alpha.

03:08 So again, t sub 1 is causing the clockwise rotation...

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