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

step 1 We're given a diagram of Atwood's Machine. There's a pulley up here, and then on one side is a m

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 running over a pulley as in Figure $\mathrm{P} 8.40 .$ 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 $T_{2}$.

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 running over a pulley as in Figure $\mathrm{P} 8.40 .$ 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

Newton’s Second Law for Linear Motion

This principle states that the net force acting on an object equals the mass of the object times its acceleration (F = ma). In systems like Atwood's machine, it is applied to each mass to relate the difference in gravitational forces and tensions to the resulting linear acceleration of the masses.

Newton’s Second Law for Rotational Motion

In rotational dynamics, the net torque acting on a body equals the moment of inertia times the angular acceleration (? = I?). For a pulley with non-negligible mass, this law is crucial in linking the difference in tensions on either side of the pulley to the rotational acceleration of the pulley.

Moment of Inertia

The moment of inertia is a measure of how mass is distributed relative to an axis of rotation, impacting how easily the object can be accelerated rotationally. In problems involving pulleys with mass, knowing the moment of inertia (which depends on the geometry and mass distribution of the pulley) is essential for calculating the angular acceleration and the associated tension differences.

No-slip Condition (Constraint Relationship)

The no-slip condition implies that the string does not slip on the pulley, meaning that the linear acceleration of the masses is directly related to the angular acceleration of the pulley by the relation a = r?. This constraint is critical in ensuring that rotational and translational motions are correctly connected in the system’s analysis.

Tension Difference in Rotational Systems

In systems where a rope runs over a massive pulley, the tensions on either side of the pulley are not necessarily equal. The difference in tension is what generates a net torque on the pulley, which is required to produce its angular acceleration. Recognizing this difference is key to accurately applying Newton’s laws in both the linear and rotational contexts.

Transcript

00:01 We're given a diagram of atwood's machine.

00:05 There's a pulley up here, and then on one side is a mass, and on the other side is another mass.

00:19 Don't know why one of them is a square, and the other one's a circle, but whatever.

00:26 And they both undergo the same acceleration.

00:32 All right.

00:35 So we're told that m1 is 10, m2 is 20, and da -da -da -da -da -da -da.

00:51 Oh, wait.

00:53 The pulley has a mass of 8, and it has a radius of 2 -tenths of a meter.

01:13 Since it is a solid cylinder, then i is going to be one half mr squared, except that was capital m.

01:32 Okay.

01:35 Let's see what we're going to do with this.

01:41 Why must t2 be greater than t1? why didn't even write t2 and t1? but nevertheless, t2 is here and here.

01:56 And here.

01:59 T1 is here and here.

02:06 Why does t2 have to be greater than t1? well, probably because m2 is much greater than m1.

02:29 It's probably the main reason.

02:34 Oh, no, it's not.

02:37 And there's inertia.

02:47 All right, b, what's the acceleration of the system? so, drawing some free doughty biograms.

03:08 For one of them, we've got m1g and we've got t1.

03:15 For the other one, we've got m2g and t2.

03:24 For the pulley, we've got t1, nope, t2 and t1.

03:36 All right.

03:38 So, some of the forces equals mass times acceleration, whoop, m1a, which is going to be t1 minus m1g.

03:57 Over here, some of the forces equals mass times acceleration, m2.

04:04 Is m2g minus t2.

04:09 Now the net torque equals i alpha.

04:19 T2 minus t1 times distance, which would be for the pulley r equals i, which is one -half, m r squared times alpha.

04:41 But also a equals alpha r so what am i trying to solve for? a, i'm trying to solve for a so alpha is a over r so i'm going to write this as 1 half m ra all right now, from here t1 equals m1a plus g.

05:33 From here, t2 is m2g minus a.

05:43 T2, it's going to be m2g minus a.

05:46 Good.

05:47 M1.

05:50 Okay, good.

05:51 So now i'll substitute that into the center equation.

05:59 T2, which is m2, g minus a, minus t1, which is m1, a plus g, i'm going to cancel out the r's here, equals one half capital m, lowercase a...

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