Vertical Circular Motion and Tension I stumbled across this problem on a physics test, and I'm convinced the correct answer (C) is wrong.

The explanation simply states that all the signs are correct, and in circular motion, inward forces are positive and outward forces are negative.
However, I believe the answer should be D. Here's my reasoning:
Tension at the highest point is zero when the centripetal force is equal to the force of gravity.
So $$mg=m\frac{v^2}r$$
and $$v=\sqrt{gr}$$
which in this case means $$v=\sqrt{9.8*0.5} = 2.21 \frac{m}s$$
Since the speed of $2 \frac{m}s < 2.21 \frac{m}s$, D should be correct because the mass would be too slow to ever complete a vertical circle.
Did I overlook something, or did the test makers make a mistake?
 A: I believe that free fall is not involved.  They are talking about forcing the mass to go 2 m/s regardless at to whether is fast enough to complete the loop in free fall.  Consider it to be a rigid rod spinning that mass.  The rod will indeed be in compression at the top of the loop because you are right that the speed is below that necessary to complete the loop in free fall.  But answer C is correct because there is nothing wrong with the calculation.
Answer D is clearly wrong, because 


*

*There can be no constant velocity vertical circular motion unless it's driven.

*For driven motion, there is no velocity too small( or too large) for the motion to be circular.
It really doesn't matter whether you choose up or down as positive.
Choose positive up and you get:
$\frac{m\ v^2}{r}-m\ g-T=0$
Go the other way you get:
$-\frac{m\ v^2}{r}+m\ g+T=0$
You get the same negative value for $T$ either way, because the rod is in comprssion.
A: I write this as an answer to explain why both of the given answers are incorrect and the OP is correct.  
The point of Physics here is that when one is using a relationship involving vector quantities  it is best to write a vector equation after defining the appropriate unit vector(s) / positive direction(s).
Choosing a positive direction is equivalent to choosing a unit vector in that direction.
In this example there a choice between a unit vector in the down direction $\hat d$ and a unit vector in the up direction $\hat u$ with $\hat d = -(-\hat d) = -\hat u$.  
It is given that the magnitude of the gravitational field strength is $g=9.8 \,\rm N\,kg^{-1}$ so as a vector one would write $\vec g = 9.8 \,\hat d$ or $9.8 \,(-\hat u)$ or $-9.8 \,\hat u$ with the $-9.8$ being the component of the gravitational field strength in the $\hat u$ direction.  
Given that it is known that in this problem the gravitational field strength and the centripetal acceleration $a_{\rm c} = 8 \,\rm m\,s^{-2}$ are in the downward direction this has been chosen to be the positive direction.
Applying Newton's second law gives $$\vec F_{\rm net} =  T \,\hat d + mg \,\hat d = m a_{\rm c} \,\hat d \Rightarrow T  + mg  = m a_{\rm c}$$ where $T$ is the component (ie it can be either positive or negative) of the tension in the $\hat d$ direction.  
On doing the sums an answer for the component of the tension in the $\hat d$ direction is obtained  $T = -1.8 m\,\rm N$.  
So $\vec T = -1.8\,\hat d$ or $1.8 \,(-\hat d)$ or $1.8 \,\hat u$ which is completely consistent with choosing $\hat u$ as the positive direction at the start and applying Newton's second law to get the component of the tension in the $\hat u$ direction.  
$$\vec F_{\rm net} =  T \,\hat u + m(-g \,\hat u) = m (-a_{\rm c} \,\hat u) \Rightarrow T  - mg  = -m a_{\rm c}\Rightarrow T = 1.8m$$ 

In the student's solution the first line indicates to me that they chose down as the positive direction which is reinforced by the fact that the numeric values for both the gravitational field strength and the centripetal acceleration were both positive when substituted into their equation.  
Looking at the options I think that the OP is correct.
