# Motion around a circle

1. Aug 3, 2015

### kaspis245

1. The problem statement, all variables and given/known data
Find the speed which one needs to impart on an object which is hanging on inelastic spring for it to make one revolution.

2. Relevant equations
Newton's Laws of Motion

3. The attempt at a solution
I have two answers for this problem, can't figure out which one is correct.

Last edited: Aug 3, 2015
2. Aug 3, 2015

### tommyxu3

The meaning of $v$ in your two equations seem different?

3. Aug 3, 2015

### tommyxu3

Here I have two opinions about your equations if the $v$ you meant is when the object is at its lowest point.
For 1), the relation $T=mg$ is not correct, for the object is moving on a circle so its net force shouldn't be $0$ but its centripetal force.
For 2), at the highest point the object is not necessarily static. To be honest, I think I should not be, for it has to keep the circular orbit it is on.

If the $v$ in your equations means the velocity at the highest point, then I think it's correct. Then you can use it to calculate the velocity of the lowest point, of course based on the conservation of the mechanic energy. Obviously, the kinetic energy of both positions shouldn't be $0.$

4. Aug 3, 2015

### kaspis245

I see. If $T$ is 0 at the highest point the object will fall, so:
$F_c=mg+T$
$\frac{mv^2}{r}=mg+ma$
$v=\sqrt{r(g+a)}$

Is this correct? What does $a$ mean?

5. Aug 3, 2015

### tommyxu3

The $v$ in your equation means the condition of the highest position? If so, then that's correct. There, $T$ can be $0,$ for there still exists gravitational force to provide centripetal force.

6. Aug 3, 2015

### kaspis245

What do you mean "$T$ can be 0"? Doesn't it mean that $v=\sqrt{gr}$?

7. Aug 3, 2015

### tommyxu3

Yes, it could be! That's what you get: The minimal speed of the object when it is at its highest position of the vertical circular motion is $\sqrt{gr}.$

8. Aug 3, 2015

### kaspis245

But that's exactly what I wrote in my first post

9. Aug 3, 2015

### tommyxu3

Yes, so in my previous posts I asked what the $v$ means in your equations, at highest or lowest points. If both of the equations mean the highest condition, then your second equation doesn't make sense.
(I'm not sure if you want to calculate the lowest position's velocity by the result you get in equation (1).)

10. Aug 3, 2015

### kaspis245

$v$ is the speed that one needs to impart on a body which is hanging on inelastic spring (at the beginning the body is at the bottom of the circle).

11. Aug 3, 2015

### tommyxu3

Ok, then clearly the first equation is wrong, for the reason I mention previously: the object is moving on a circle so its net force shouldn't be 0 but its centripetal force.
Your problem in the second one is also mentioned in the same post. The kinetic energy when the object reach the highest must not $0,$ so there may lack something in your equation:
$$\frac{1}{2}mv^2=mg(2r)+\text{the minimal kinetic energy it has at the highest}.$$
And, in our discussion above, we have got the minimal speed it needs at the highest, from the equation $mg=m\frac{v_{high}^2}{r},$ which gives us $v_{high}=\sqrt{gr}$.
So, you just have to solve this:
$$\frac{1}{2}mv^2=mg(2r)+\frac{1}{2}mv_{high}^2.$$

12. Aug 3, 2015

### kaspis245

Ok, now everything seems clear. So the answer is $v=\sqrt{5gr}$.

13. Aug 3, 2015

### tommyxu3

That's right! Great!