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An object mass $1 \, \rm kg$ hangs on a rope length $1 \, \rm m$. The object gets pushed with minimal horizontal velocity required to swing a full vertical circle, i.e.: keeping the rope stretched.

What time $t_c$ does it take to complete that intact circle?

I added conclusion near end.

I have seen similar questions and answers to describe minimal horizontal velocity and variable velocity along the circular swing path, and to compute maximal tension on the rope. They all use conservation of kinetic + potential energy and balance centrifugal force with gravitational force at the top of the circular swing path.

But I have not seen the time described.

I ask because I wonder how that time compares to time $t_v$ when just pushing the object straight upward with minimal vertical velocity required to reach same top height $2 \, \rm m$ and to come down again.

Because in the first case the minimal velocity is bigger than in the second case, but, in the second case the distance is smaller than in the first case.

So would $t_c$ be bigger or smaller or equal to $t_v$ ?

I found (classical exercise: using distance formula for falling object (no rope attached) in time)

$t_v \simeq 1.3 \, \rm s$

and

$t_c \simeq 1.2 \, \rm s$

but I am not sure I am reasoning correctly.

On the other hand, times near $1 \, \rm s$ feel realistic to me for both.

As requested: here is my original approach which I was not sure to be correct.

>>> import numpy as np
>>> from math import pi, sqrt

# define velocity in function of height
# see https://www.geeksforgeeks.org/motion-in-a-vertical-circle/
>>> def v(h):
...     return sqrt(9.81*(5-2*h))
... 

# test some values
>>> v(0)
7.003570517957251
>>> v(1)
5.424942396007538
>>> v(2)
3.132091952673165

# f(h) is the (non elliptic) integral of v(h) (Wolfram)
>>> def f(h):
...     return -1.04403*(5-2*h)**(3/2)
... 

# find average velocity
>>> (f(2)-f(0))/2
5.314290126372764

# find time
>>> (2*pi)/5.314290126372764
1.182318834268865

Later, having seen expert answers I replaced height by angle.

code:

# import some stuff

>>> import numpy as np
>>> from scipy.integrate import quad
>>> from math import pi, sin, sqrt

# velocity function v
# note h = 1 - sin(x) and v^2 = g * (5 - 2 * h)

>>> def v(x):
...     return sqrt((9.81)*(3-2*sin(x)))
... 

# test some values

>>> v(-pi/2)
7.003570517957251
>>> v(0)
5.424942396007538
>>> v(pi/2)
3.132091952673165

# values look as expected

# integrate velocity from -pi/2 to +pi/2

>>> res, err = quad(v, -pi/2, pi/2)
>>> print(res)
16.507274579464244


# find average meters per second

>>> res/pi
5.254428692593845

# find time for distance 2pi

>>> print((2*pi)/(res/pi))
1.1957884814453341

Note that in both cases $t_c \simeq 1.2$

I am also not sure how precise numerical integration really is.

So I asked question in this exchange to compare outcome.

conclusion: mistake in my approach is to calculate average velocity to find time. One must integrate inverse of angle velocity to find time.

Both given answers are correct. They give $t_c \simeq 1.289 s$ and so $t_c$ is a little bit bigger than $t_v \simeq 1.277 s$.

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  • $\begingroup$ Thanks for adding homework-and-exercises tag. I am new here and not really sure if it is biased rather positive or negative, but, overall, I am happy to get some as opposed to none reaction. $\endgroup$ Feb 23, 2023 at 2:08
  • $\begingroup$ How did you calculate the time $t_c$? Did you integrate the velocity around the circle? $\endgroup$
    – BowlOfRed
    Feb 23, 2023 at 3:10
  • $\begingroup$ @BowlOfRed like gandalf61 I integrated the velocity around the circle, but, not the angular velocity, rather the length of the tangent velocity vector. Perhaps that's a wrong approach. $\endgroup$ Feb 23, 2023 at 16:31
  • $\begingroup$ The time to push the bob and almost reach the top dead center and come down again can be almost infinite depending on how close the vertical the rope goes. $\endgroup$ Feb 24, 2023 at 15:11
  • $\begingroup$ @JohnAlexiou : for $t_v$ I was not thinking about having a rope attached. I should have mentioned that. $\endgroup$ Feb 24, 2023 at 17:48

2 Answers 2

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enter image description here

the energy is:

$$E=\frac 12\,L^2\,m\,\dot\phi^2-m\,g\,L\,\cos(\phi)$$

and the rope tension T :

$$T=-m\,(\dot\phi^2\,L+g\,\cos(\phi))$$

at t=0

$$E_0=\frac 12\,L^2\,m\,\dot\phi^2(0)-m\,g\,L\,\cos(0)$$

the rope tension at $~\phi=\pi~$ must be greater then zero

$$T(\phi=\pi)\ge 0=-m\,(\dot\phi^2(\pi)\,L+g\,\cos(\pi))\quad \Rightarrow\\ \dot\phi^2(\pi)=\frac gL\quad, \dot\phi^2(0)=-\frac gL$$

from here solving $~E=E_0~$ for $~\dot\phi~$ you obtain

$$\frac{d\phi}{dt}=\sqrt{\frac gL}\sqrt{3-2\,\cos(\phi)}\quad\Rightarrow\\ \int_0^{2\,\pi} \left(\sqrt{\frac gL}\sqrt{3-2\,\cos(\phi)}\right)^{-1}\,d\phi=\int_0^{t_f}\,dt$$

this is Elliptic Intergral , you obtain $~t_f~\approx 1.29\,[s]$

the start velocity $~v_0=L\,\dot\phi(0)=-L\,\sqrt{g/L}~$

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  • $\begingroup$ Do you have a few more digits or is it exactly $1.28$? I ask because for $t_v$ I find $1.277101713628202$. So which one ($t_v$ or $t_c$) would be larger? $\endgroup$ Feb 23, 2023 at 21:04
  • $\begingroup$ Having $L=1$ isn't $\sqrt{g} \simeq 3.13$ the velocity for $\phi = \pi$ at the top? $\endgroup$ Feb 24, 2023 at 0:23
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    $\begingroup$ Probably it should be $ ~3-2\cos(\varphi)~$ from 0 to $2\pi$ $\endgroup$
    – Eli
    Feb 24, 2023 at 18:14
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    $\begingroup$ right: and the time is $1.289 s$ $\endgroup$ Feb 24, 2023 at 19:37
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    $\begingroup$ the solution is $~4.037811638\,\sqrt{\frac lg}~$ and g=9.81.... $\endgroup$
    – Eli
    Feb 24, 2023 at 22:17
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Let $\omega(\theta)$ be the angular velocity of the object when the taut rope makes an angle $\theta$ with the horizontal (taking angles below the horizontal to be positive). By conservation of energy we have

$\displaystyle \frac 1 2 m r^2 \omega(\theta)^2 = \frac 1 2 m r^2 \omega(0)^2 + mg r\sin(\theta) \\ \displaystyle \Rightarrow \omega(\theta)^2 = \omega(0)^2 + \frac {2g} r \sin(\theta)$

To keep the rope taut the tension in the rope must be greater than or equal to zero throughout the circle. In particular, at the top of the circle we must have

$\displaystyle T = mr \omega\left(-\frac \pi 2 \right)^2 - mg \ge 0 \\ \displaystyle \Rightarrow \omega\left(-\frac \pi 2 \right)^2 \ge \frac g r \\ \displaystyle \Rightarrow \omega(0)^2 \ge \frac {3g} r$

To maximise the time taken to complete a circle we must minimise $\omega(\theta)$ and so we must minimise $\omega(0)$. So for maximum time we have

$\displaystyle \omega(0)^2 = \frac {3g} r$

and so

$\displaystyle \omega(\theta)^2 = \frac g r (3+2 \sin(\theta) ) \\ \displaystyle \Rightarrow \omega(\theta) = \sqrt {\frac g r (3+2 \sin(\theta) )}$

Now that you have an expression for $\omega(\theta)$ you can in principle find the time to complete half a circle, which is

$\displaystyle \int _{-\frac \pi 2} ^{\frac \pi 2} \frac 1 {\omega(\theta)} d \theta$

and then you just double this to find the time to complete a full circle. In practice, I think finding an exact expression for the integral may be difficult.

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  • $\begingroup$ Thanks for answer. So, am I right to calculate required minimum velocity at lowest point is around $5.4$ meters per second? I ask because elsewhere I found around $7$ meters per second. Sorry to miss the ball. $\endgroup$ Feb 23, 2023 at 16:36
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    $\begingroup$ @FirstNameLastName The minimum velocity at the lowest point, when $\theta = \frac \pi 2$, is $\sqrt {5gr}$. If $r$ is 1 metre then $\sqrt {5gr}$ is about 7 m/s. $\endgroup$
    – gandalf61
    Feb 23, 2023 at 17:06
  • $\begingroup$ Right, I got your angle 90 degrees wrong. If $v$ is velocity at height $h$ (taking $h=0$ at lowest point), $v$ is indeed 7 and 5 halfway and 3 on top (approximately). And $v^2 = 5g-2h$. I then integrated $v$ in function of $h$ from $0$ to $2$ to help find mean value to use to calculate time. But I am not sure if this method is valid or if my calculations are right. So I wondered if anyone has exact value for the time. $\endgroup$ Feb 23, 2023 at 18:00
  • $\begingroup$ sorry it is : $v^2 = g(5 - 2h)$ which I found on geeksforgeeks.org/motion-in-a-vertical-circle $\endgroup$ Feb 23, 2023 at 18:34
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    $\begingroup$ @FirstNameLastName Eli has gone further than I did and actually evaluated the integral, so they should have the tick. $\endgroup$
    – gandalf61
    Feb 24, 2023 at 20:46

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