Solving for Tension in Net Force Equation: Need Help!

[Dun Artorias]

Homework Statement

A wheel is mounted in a physics lab as shown above. The wheel has a mass M = 2.2 kg and radius R = 35 cm. The wheel is in the shape of a ring with rotational inertia I = MR^2. A object of mass m = 5 kg is attached to the end of a string which is wrapped around an inner hub (on the left hand side) with radius r = 2.7 cm. The object is released from rest a distance h = 77 cm above the floor and begins to move downward.

Determine the linear acceleration of the object.

Relevant Equations

Net Force = ma
Net Torque = I * alpha

I have tried finding the tension of the string through the equation
Net Force = T - mg = ma
but I am struggling with coming up with the correct acceleration. Is my net force equation correct?

 

[PeroK]

Can you include a diagram for the problem?

 

[haruspex]

Is my net force equation correct?

It depends which way you are defining as positive for the acceleration.
What about a corresponding equation for the wheel?

 

[erobz]

I'm finding that the hanging mass will experience a constant linear acceleration less than $g$ If I've interpreted it correctly.

 

[kuruman]

I agree with @haruspex that your equation is ambiguous but not because it depends on whether you define the acceleration positive when it is up or positive when it is down. It's because, apparently, you didn't incorporate the known direction of the acceleration in your equation. See the explanation below which is lengthier than originally planned and I apologize for that.

Here we have a 1 D vector equation expressing Newton's second law. Using arrows over symbols to indicate vectors, $$\sum_i \vec{F}_i=m\vec a.$$Now $$\vec T+m\vec g=m\vec a$$Note that a "plus" sign appears between all vectors because the net force is the sum of all forces in Newton's law. In 1D, the equation is simplified by removing the arrows to convert the vector equation into a scalar equation. There are two possibilities.

Case I: You know the directions of all vectors in the 1D equation.
First you pick a direction as positive. Second, you replace the arrows with "+" sign in front of the vector symbol if it points along the chosen positive direction and a "-" sign if the vector points opposite to the chosen positive direction. In this particular problem you know that the direction of the tension is "up", the direction of the weight is "down" and the direction of the acceleration is also "down." Following this rule,

1. If you choose "up" as positive $~\vec T+m\vec g=m\vec a \rightarrow T-mg=-ma.$
2. If you choose "down" as positive $~\vec T+m\vec g=m\vec a \rightarrow -T+mg=ma.$

Note that one equation becomes the other if you multiply both sides by $-1$ which is equivalent to choosing the opposite direction as positive. Also, symbols $T$, $g$ and $a$ in the equations stand for magnitudes of vectors and are positive quantities. If you replace symbols with given numbers in either equation and the acceleration comes out negative, you will know that you made a mistake somewhere. The magnitude of a vector is positive independently of the choice of axes.

Case II: You don't know the direction of one vector in the 1D equation.
You follow the same rule with the additional directive "keep it simple". We can modify the situation in this problem a bit and say the we have a vertically accelerating mass $m$ attached to a string in which the tension is $T$. We are asked to find the acceleration, a vector, in terms of the other quantities.

First note that the tension is "up" and the weight are "down", therefore the tension and the weight must appear with a relative negative sign in the expression for the net force regardless of whether "up" is positive or negative. Newton's second law can be written in vector equation form (a bit unconventionally but clearly as to what is meant) $$\frac{1}{m}T\text{(up)} +g\text{(down)}=a\text{(?)}$$ We can remove the directions that we know and replace them with negative signs. For the acceleration, we "keep it simple" and assume that the unknown direction denoted by the question mark is positive as defined by the choice of positive. Note that here $a$ stands for the scalar component of the vector not its magnitude; it could be positive or it could be negative, we just don't know. We get

1. $a=\dfrac{1}{m}T -g~$ if "up" is positive
2. $a=-\dfrac{1}{m}T +g~$ if "down" is positive

Either equation is correct and equation 1 is the form that you have posted. Note that one does not get one equation from the other if both sides are multiplied by $-1$, i.e. they are not the same equation. That is, I think, what @haruspex meant when he said

It depends which way you are defining as positive for the acceleration.

Nevertheless, if you substitute numbers in either equation and the acceleration turns out to be a negative number, it will mean that the initial assumption (when the question mark was removed) was incorrect. The acceleration must be opposite to the chosen positive direction. The catch here is that if you make a calculation error and the acceleration comes out with one sign when it should be the other, you wouldn't know it. That is why if you know the direction of the acceleration, you should put it in as shown in case I.

 

[haruspex]

Since @Dun Artorias wrote T-mg, it is clear that up is being taken as positive for T. The remaining question is whether up is also positive for the acceleration.

Case I: You know the directions of all vectors in the 1D equation.
First you pick a direction as positive.

1. If you choose "up" as positive $~\vec T+m\vec g=m\vec a \rightarrow T-mg=-ma.$
2. If you choose "down" as positive $~\vec T+m\vec g=m\vec a \rightarrow -T+mg=ma.$

I don't understand that. If |T|>mg and up is chosen as positive (1) will give a negative a, meaning it accelerates down?
Guessing correctly which way things will move is of no benefit. The equations are the same.

With up positive for everything, T-mg=ma.

With down positive for everything, it depends whether you think of T as the magnitude of the tension in the string section or as the scalar value of the force exerted on the mass by that tension.
In the magnitude view, -T+mg=ma; in the scalar view, T+mg=ma, with T being negative.

 

[Lnewqban]

Homework Statement:: A wheel is mounted in a physics lab as shown above.

 

[erobz]

I wouldn’t hold your breath, you’re turning blue.

 

[Lnewqban]

I wouldn’t hold your breath, you’re turning blue.

I see a vectorial discussion developing on one unique post and a promised picture that never made it into this thread.
Please, see post #2.

 

[erobz]

I see a vectorial discussion developing on one unique post and a promised picture that never made it into this thread.
Please, see post #2.

I was trying to make a joke about the color of your emojis…I’m afraid it didn’t land.

 

[Lnewqban]

I was trying to make a joke about the color of your emojis…I’m afraid it didn’t land.

Oh no, please; I got the joke, and I liked it.
My comment was not on that.

Perhaps others can, but I can't visualize the problem's set up only from the OP's description.
A posted diagram, as requested in post #2, would be nice.

 

[erobz]

Oh no, please; I got the joke, and I liked it.
My comment was not on that.

Perhaps others can, but I can't visualize the problem's set up only from the OP's description.
A posted diagram would be nice.

I picture wheel fixed to the wall on an axle, and a string wrapped around a small hub on the axle with the weight attached to it dropping to the floor.

A posted diagram would be nice, but the OP is MIA.

 

[Lnewqban]

I believe that I have finally seen it: