Newton's 2nd Law for Rotation

In summary, the problem involves a uniform disk with a block hanging from a massless cord that is wrapped around the rim of the disk. The cord does not slip and there is no friction. Using the equations for net torque and tangential acceleration, the acceleration of the block is found to be -4.8 m/s^2. However, this answer is incorrect due to a mistake in the model for the net force. The net force is always equal to ma, with both having the same sign. The reason for using -TR in the net torque equation is because T is defined as the force on the hanging mass, not the force on the disk. Therefore, the correct equation for the net force is T = ma.
  • #1
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.
 
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  • #2
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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What is Newton's 2nd Law for Rotation?

Newton's 2nd Law for Rotation states that the angular acceleration of an object is directly proportional to the net torque acting on it and inversely proportional to its moment of inertia. In simpler terms, it explains the relationship between an object's rotational acceleration, the forces acting on it, and its resistance to rotational motion.

What is the formula for Newton's 2nd Law for Rotation?

The formula for Newton's 2nd Law for Rotation is τ = Iα, where τ represents the net torque, I is the moment of inertia, and α is the angular acceleration.

How is Newton's 2nd Law for Rotation different from Newton's 2nd Law for linear motion?

Newton's 2nd Law for linear motion deals with the relationship between an object's mass, acceleration, and the force applied to it. On the other hand, Newton's 2nd Law for Rotation explains the relationship between an object's rotational acceleration, torque, and moment of inertia.

What is moment of inertia?

Moment of inertia is a measure of an object's resistance to rotational motion. It is dependent on an object's mass and how it is distributed around its axis of rotation. Objects with a higher moment of inertia require more torque to achieve the same angular acceleration compared to objects with a lower moment of inertia.

How can Newton's 2nd Law for Rotation be applied in real-life situations?

Newton's 2nd Law for Rotation can be applied in various real-life situations, such as understanding the movement of objects like cars and bicycles, analyzing the performance of machines like engines and turbines, and designing structures like bridges and buildings to withstand rotational forces. It is also essential in sports like gymnastics and figure skating, where rotational motion is a crucial component.

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