How Does Newton's Second Law Apply to Rotational Motion in a Pulley System?

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SUMMARY

The discussion centers on applying Newton's Second Law to a pulley system involving a uniform disk and a hanging block. The system is defined by a disk with a mass of 2.5 kg and a radius of 0.20 m, with a block of 1.2 kg suspended from a massless cord. The net torque and force equations are derived, leading to the correct acceleration of the block being calculated as -4.8 m/s². The confusion arises from the treatment of signs in the equations for net force and torque, emphasizing the importance of consistent directional definitions in physics.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with rotational dynamics (τ = Iα)
  • Knowledge of torque and its directionality in physics
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the relationship between linear and angular acceleration in rotational systems
  • Learn about the moment of inertia for different shapes (e.g., disks, spheres)
  • Explore the concept of tension in pulley systems and its effects on motion
  • Investigate common mistakes in applying Newton's laws in multi-body systems
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Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for clarification on common misconceptions in applying Newton's laws.

rainstom07
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Homework Statement


In the figure http://i.imgur.com/Y5Vc7.gif shows a uniform disk (M = 2.5kg, R = 0.20 m) mounted on a fixed horizontal axle. A block (m = 1.2kg) hangs from a massless cord that is wrapped around the rim of the disk. The cord does not slip and there is no friction.

Find the acceleration of the block.

Homework Equations


[tex]\tau_{net} = I\alpha[/tex]
[tex]a_t = \alpha r[/tex]

The Attempt at a Solution


1. [tex]F_{net} = -ma = T - mg[/tex]
2. [tex]a = -\frac{T}{m} + g[/tex]

3. [tex]\tau_{net} = I\alpha = \frac{1}{2}MR^2\alpha = -TR[/tex]
4. [tex]\alpha = \frac{a_t}{r} = \frac{a}{r}[/tex] the tangential acceleration is also the linear acceleration.

5. [tex]-TR = \frac{1}{2}MR^2(\frac{a}{r})[/tex]
6. [tex]T = -\frac{1}{2}Ma[/tex]

Plugging equation 6 into 2 yields:
[tex]a = \frac{Ma}{2m}+g[/tex]
Some manipulation

[tex]a = \frac{g}{1-\frac{M}{2m}}[/tex]
[tex]a = \frac{9.8}{1-\frac{2.5}{2*1.2}} = -235.2[/tex]

This is obviously wrong.

5. Solution
7. [tex]F_{net} = ma = T - mg[/tex]
8. [tex]a = \frac{T}{m} - g[/tex]

plugging equation 6 into equation 8 yields:
[tex]a = -\frac{g}{1+\frac{M}{2m}}[/tex]
[tex]a = -\frac{g}{1+\frac{2.5}{2*1.2}} = -4.8 ms[/tex]

6. My question
Obviously my model for the net force is wrong. I used -ma instead of ma.

Why is it okay for me to consider the direction of the net torque to be -TR instead of TR for equation 3, but it isn't okay for me to consider the direction of the net force to be -ma instead of ma for equation 1? The net force is pointing downward isn't it?

Also vice-versa
Why is it okay for me to consider the direction of the net force to be ma instead of -ma for equation 1, but it isn't okay to consider the direction of the net torque TR instead of -TR for equation 3? shouldn't i be consistent?

Thanks in advance. This stuff always gets me in physics.
 
Last edited:
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rainstom07 said:
6. My question
Obviously my model for the net force is wrong. I used -ma instead of ma.

Why is it okay for me to consider the direction of the net torque to be -TR instead of TR for equation 3, but it isn't okay for me to consider the direction of the net force to be -ma instead of ma for equation 1? The net force is pointing downward isn't it?

Also vice-versa
Why is it okay for me to consider the direction of the net force to be ma instead of -ma for equation 1, but it isn't okay to consider the direction of the net torque TR instead of -TR for equation 3? shouldn't i be consistent?

Thanks in advance. This stuff always gets me in physics.

F is always equal to ma. They are vectors, which means they have both magnitude and direction. In this 1D problem, this translates to them being either negative or positive, depending on whether they point downward or upward. But they must both have *same* sign, regardless of what it is. If you write F = -ma, then what you are saying is that the net force points in the opposite direction of the acceleration! I.e. if the net force is upward the acceleration is downward, and vice versa. This is clearly not the case.

What's tricky here is that the sign here is intrinsic i.e. it is not indicated explicitly. F and a are numbers that could either be positive or negative e.g. F = +10 N, a = +4 m/s^2, or F = -10 N, a = -4m/s^2. So if the net force is pointing downward, then that will be reflected in your numerical answer. But the reason you don't put the negative sign in explicitly is because F and a are assumed to be signed quantities that carry that sign already.

EDIT: the reason why it was okay for you to write -TR was because you defined T as the force *on the hanging mass* due to the tension in the rope. The force *on the disc* due to the tension in the rope will necessarily be -T then. It's a different force.
 
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