I am thinking of comparing horizontal to vertical circular motion. I think I want to compare (both experimentally and mathematically) the constant tension of horizontal to the varying tension of the vertical circular motion.

Here is a concept idea of the experiment:

enter image description here It is the same as your generic high school circular motion experiment except for this time I hand spin the mass vertically instead of horizontally.

So my questions: Is the slotted mass responsible for the centripetal force in vertical circular motion? How can I calculate the initial velocity of the mass? Given the small mass ($25\ \mathrm g$), the mass of the slotted mass ($150\ \mathrm g$), radius ($1\ \mathrm m)$ and period ($0.7$) available.

Also is this a good experiment to do (constant tension vs varying tension)? What variables should I change instead?

  • $\begingroup$ Could you share what the 'generic high school' horizontal circular motion experiment looks like? I do not know as your and my experimental curriculum seems to differ. $\endgroup$ Mar 26, 2020 at 9:34
  • $\begingroup$ See Two types of tension equation in vertical circular motion (confusion) $\endgroup$ Mar 26, 2020 at 11:47
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    $\begingroup$ Sorry, but your question is mostly off-topic here, as it contravenes our policy regarding homework-like questions. OTOH, this part is ok (IMHO): "Is the slotted mass responsible for the centripetal force in vertical circular motion?" because it is asking a conceptual question. $\endgroup$
    – PM 2Ring
    Mar 29, 2020 at 16:35
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    $\begingroup$ @Negrawh Following PM2Ring's comment, I would shift your question away from the subjective question of "is this feasible" / "what variables should I change" and focus on the physics concepts you want to learn more about $\endgroup$ Mar 29, 2020 at 18:41
  • $\begingroup$ That should be ok, as long as you do make it conceptual, and avoid the calculation details. You can also ask about what part your hand plays in all this. After all, there won't be circular motion if your hand does nothing. ;) You can even link to this question, if you like, and mention that it's a follow-on question, and explain what stuff you need that these answers don't cover. That may stop people from voting to close your new question as a dupe of this one. $\endgroup$
    – PM 2Ring
    Apr 3, 2020 at 5:59

2 Answers 2


The biggest problem will be making any practical measurement with the vertical plane rotation. The tension in the string will no longer be constant, as you clearly understand, but that means that the hanging slotted weight (m) will never balance the tension (well, actually it will 'balance' at two specific angles, bjut that is like saying that a stopped clock is correct twice a day).

So during each 2Pi rotation the tension will either be greater than mg, so the weight will accelerate upwards, or will be less than mg, and the weight will accelerate downwards. Measuring the motion of a weight where it's acceleration is varying continuously will be difficult if at all practical. In fact it isn't clear to me that you will have any sort of equilibrium (as you do in the horizontal experiment) - I believe that it is likely that at any constant rotation rate the hanging mass will either drop to the floor or will accelerate upwards until it hits the 'blue' box in your diagram.

In practice, it might be better to replace the hanging mass with a spring balance where, with care you could at least measure the maximum tension in the string as a function of rotational velocity and radius of rotation. Alternately, if you want to keep the hanging mass (or if that is all you have to use) one nice experiment is to drop the rotating mass from various heights (and with varying string length) to see at combinations of drop height and arc radius provide sufficient tension to 'just' lift the slotted weight as the rotating mass reaches the bottom of its arc. (see sketch below). Of course that makes it a very different experiment to the horizontal rotating mass experiment.

enter image description here

Just calculating the dynamics of the vertical rotating system (with hanging mass) might have some fairly challenging maths involved. But I think you have the right research spirit to go a long way at whatever you try.

  • $\begingroup$ Hey man thanks for the reply! I actually did the experiment, and I was able to maintain rotation and collected period. And for your suggested modification, I am afraid to say the task require us to relate with circular motion $\endgroup$ Mar 26, 2020 at 12:00
  • $\begingroup$ Penguino does the slotted mass in vertical motion still indicates the tension force? What is the purpose of the slotted mass in the vertical case $\endgroup$ Mar 27, 2020 at 1:55
  • $\begingroup$ @Negrawh if I understand your exp. setup, you rotate the red mass M in a vertical circle, attached to a string that can slide freely through, the blue component, and is attached to the hanging slotted mass m. If that is the case then tension T in the radial part of the string due to rotating M equals tension in the lower string. So the slotted mass will feel upward force T. If mg>T, the small mass will have acceleration a=mg-T downwards, if mg<T then a will be upwards. mg=T then a=0. Problem is, a varies through rotation so m will likely have a, v, and height h all complicated functions of t. $\endgroup$
    – Penguino
    Mar 28, 2020 at 2:31
  • $\begingroup$ @Negrawh As above, motion of m may be quite complex and it will be difficult in general to measure when a is +ve or -ve. The exception to that rule is if T < mg throughout almost all the cycle of the mass M, m is stationary and resting on a surface, and the string is tight but not lifting m off the surface. Then if, for example, you increase rotational vel. of M, you will be able to see the point at which T rises above mg - as m will immediately be accelerated upwards and will jump up off the table. But I don't think there is a steady state equiv of that seen in horizontal rotation expt. $\endgroup$
    – Penguino
    Mar 28, 2020 at 2:38

From the diagram below, how does the newton's 2nd law equation change from top to bottom of the motion? (with mv^2, r and T). Whilst the slotted mass's weight will indicate the tension, the overall resultant force changes because of the mass of the object you are rotating is either added or taken away from the tension.

You should have the time period for a specific number of rotations, when you measure this, you can then caculate the angular velocity. Then you should be able to move on to calculating the velocity.

Exactly, what is the scientific question you are trying to address? You should try to write in in the format

How does x affect y when a,b and c are constant?

For example "How does the tension on a string affect the angular velocity of a rubber bung in circular motion if the radius is kept constant?"

"How does vertical circular motion of a rubber bung affect the centripetal force acting on it if the length of the string is kept constant?"

Once you have an idea of this, you can build a mathematical model, put some equations to it, take you measurements, plot them on a suitable graph and then develop a conclusion and evaluation.

I am speaking as an A Level UK teacher.

forces in vertical circular motion

  • $\begingroup$ Weight of slotted masses = tension of the string $\endgroup$ Mar 29, 2020 at 14:57