# B Why is there "weightlessness" on the top of a verticle circle?

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1. Mar 7, 2016

### 1832vin

i'm ashamed, that i never understand this, eventhough i'm studying quantum mechanics....

so..... why is there "weightlessness" on the top of a verticle circular motion?
ie, if a plane if flying in verticle circles, why is there weightlessness while on the top of a circular path?
i mean, if it's weightlessness, that means that the wieght of the object is cancelled by a force equal and opposite, leaing to no net force, not change in interia, and therefore weightlessness....

however, that's not the case.....? on the top of a circlar motion/path, the wieght points down to earth, whilst the centripital force is also pointing downwards.... wouldn't that make the person to feel even more force applied downwards?

the answers always "the weight = centriplital force" but i don't get how does that mean "weightlessness"

2. Mar 7, 2016

### Svein

In order to stay in the circle, you need a force directed towards the center of the circle and with magnitude $C=m\cdot \frac{\omega^{2}}{R}$. At the top of the circle, you still need that force and the only force you have available is $F=m\cdot g$ which is pointed downwards. Therefore, the net "weight" at the top of the circle is $F-C=m\cdot g-\frac{\omega^{2}}{R}$. From this formula you can see that $F-C=0$ when $m\cdot g-m\cdot \frac{\omega^{2}}{R}=0$.

Last edited: Mar 7, 2016
3. Mar 7, 2016

### PeroK

In fact, if you are moving faster than the minimum speed to maintain the circular orbit, then your weight is pointing upwards when you're at the highest point.

4. Mar 7, 2016

### Staff: Mentor

Why not draw a free body diagram for an object sitting on earth and see how it differs from what you drew?