# Vertical uniform circular motion.. can it really be uniform? [closed]

This is the picture in my mind:

For centripetal force, I learned that: $T-mg\cos\theta= \frac{mv^2}{r}$

In vertical circular motion, the velocity is naturally going to decrease as kinetic energy is converted to potential energy as the particle moves up the circle- and thus resulting into a non-uniform circular motion.

However, I was told that if tension changes accordingly, the velocity will remain constant(as from the equation). What I do not get about this part of the explanation is that which force will balance mgsinθ (tangential force in the diagram )?

## closed as off-topic by Waffle's Crazy Peanut, tpg2114♦, Brandon Enright, Qmechanic♦Dec 3 '13 at 1:02

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• The equation you are giving is a condition on T so that the movement be circular. The equation does not give an explanation about the time evolution of v. You are absolutely right. There is no way that the speed can be a constant. The tension does no work on the system. Therefore, the only force entering the energy conservation is gravity -> the mass will be slower at the top. Another way to see it is as you mention in your post : nothing in this system can cancel mg sin$\theta$. – Mathusalem Dec 2 '13 at 13:46
• @Mathusalem: could you please elaborate "time evolution of v" – Eliza Dec 2 '13 at 13:55
• why close vote? – Eliza Dec 2 '13 at 16:18
• I also don't understand why this question was closed. – user55515 Mar 31 '16 at 12:12

## 2 Answers

In the situation at hand, you'll never be able to achieve uniform circular motion.

$\frac{d\vec{v}}{dt} = \frac{1}{m}\sum \vec{F}_{\text{ext}}$

This is a vectorial equation. If you look at the picture you've drawn, you have forces on the radial as well as the tangential direction.

On the radial direction, there is the tension force, and $mg\cos\theta$. The equation you have given is the outcome of a calculation which tells you that in order to have the object moving at velocity $v = |\vec{v}|$ on a circle of radius $R$, you have to apply to it a net perpendicular force of equal to $\frac{mv^2}{R}$. That's all it says.

However, there is the tangential part still.

$\frac{d}{dt}v_{tan} = mg\sin{\theta} \neq 0 \text{ in general}$

This means that the speed along the circle is always subject to a gravitational force. Thus, the speed along the circle varies. Since the tension force is orthogonal to it, it can not act on the magnitude of this speed, it can only act on its direction, so it can do nothing to keep it constant. The examples given in the reply of User58220 are misleading because they have added mechanisms which contribute to give forces such, that the tangential part of the gravitational force is cancelled.

In short : no, tension won't make it go at a constant speed.

• :so unless there is a force in the direction of mgsinθ (which tension or thrust cannot provide), the particle will just move with non-uniform circular motion? – Eliza Dec 2 '13 at 16:25
• Yes. You need a force to annihilate mgsin$\theta$. – Mathusalem Dec 2 '13 at 16:49
• In case of the London eye (example from user 58220), what is the "mechanism" that annihilates mgsinθ? – Eliza Dec 2 '13 at 16:58
• Well there is a motor making the whole structure spin. If the motor doesn't spin, neither does the London eye. – Mathusalem Dec 2 '13 at 17:34

The answer to your question is "Yes, if you want it badly enough.

Uniform vertical circular motion implies that an object is moving in a circle, that the plane of the circle is vertical, and that the speed of the object does not change as it moves around the circle. An example of such motion is that of a point on the end of the hour or minute hand of a tower clock. Another is the cars of a Ferris Wheel like the London Eye, which never changes speed; you get on and off a moving car.

It is true that gravity exerts a force that has a component that will tend to slow the object as it rise, and speed it up as it falls. So if you require uniform motion, you need a way of applying a balancing force to keep the vertical motion uniform. A rigid rod, rather than a string, can apply tension, thrust, and torque, while a string cannot...

• :In case of the London eye, why wouldn't the speed change, isn't gravity acting on the cars of the ferris wheel as they go up?Also, in case of a rod, how would it be able to exert a force to balance mgsinθ, isn't thrust along the direction of the rod (radial rather than tangential)? – Eliza Dec 2 '13 at 16:12
• The London Eye has a motor and rigid "spokes" to tranfer a tangential force to the rim; in addition, the cabins rising on one side are balanced by those descending on the other side. A rigid rod can exert a tangential force, as well as a radial pull or push. – DJohnM Dec 2 '13 at 18:30