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?

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.

• g down is larger than the centrifugal force up so the rod has to push up by an amount 1.8m to balance the load at the top. This means the rod is in compression and the tension is negative. A positive tension indicates pulling. – Bill Watts Jan 22 at 8:13
• @BillWatts I entirely agree but look towards the end of option (C) and my comment under your answer. – Farcher Jan 22 at 8:24
• See my update . – Bill Watts Jan 22 at 8:37

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

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

2. 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.

• Hmmm... But the question says "vertical circle" and also how would this explain the tension being negative? – RayDansh Jan 22 at 1:46
• Negative tension is compression. At the top of the loop the rod is pushing rather than pulling which is what compression is. You may be assuming a string pulling this mass around, but there is nothing in the problem that says that. It says the mass velocity is constant which means it is being driven. If it were in free fall, it would not be constant. – Bill Watts Jan 22 at 1:57
• Ahhh, I see. Thanks for the explanation on negative tension. – RayDansh Jan 22 at 1:59
• @Farcher I don't know if the first equation is copied wrong or not, but it at least has a typo with m a^2/2. The second equation is correct with g negative. Either way, a negative tension indicates compression and the rod is definitely in compression at the top of the loop. – Bill Watts Jan 22 at 8:03
• @BillWatts I agree there are some typos including $a^2$ instead of $v^2$ but the point I am trying to make is that the force on the mass due to something is in the upward direction.. The OP's analysis at the end of the question is correct. Think of it this way. In free fall the acceleration is 9.8 whereas all that is needed to move in a circle is an acceleration of $8$ so something must be pushing up on the mass. Option (C) is incorrect because it states that the tension comes out to be negative when up was chosen to be positive. – Farcher Jan 22 at 8:23