# Question about pendulum

I came up with this problem by myself: How much force do I need to make a pendulum revolve? Now I imagined that the force $\vec{F}$ must be enough to make the pendulum swing until half of the circumference; I tried using the formula for the period if the pendulum $T=2\pi \sqrt{\frac{l}{g}}$ and formulas from circular motion but I still had lot of problems. Can someone help me? Also consider that the pendulum is has a weight.

• I am unable to make sense out of your question. Could you try to rephrase it or otherwise put it in a clearer form? – Danu May 18 '14 at 11:16

If we assume that the angle of swing is small the solution to the equation of motion of your pendulum is:

$$x = A \sin \frac{2\pi t}{T}$$

where $T$ is the period you've given in your question and $A$ is the maximum amplitude. So let me rephrase your questions as:

If I know the period $T$ what force do I need to make the pendulum swing with amplitude $A$?

To answer this we calculate the velocity of the pendulum bob, $v$, using:

$$v = \frac{dx}{dt} = A\frac{2\pi}{T} \cos \frac{2\pi t}{T}$$

Now what we'll do is work out the momentum of the bob at the centre point, where it's moving fastest, and then we can calculate what force is required to impart that momentum.

At the centre point $x = 0$, so $t = 0, 2\pi, 4\pi$ etc, ad if we put these values for $t$ into the expression for the velocity we get:

$$v_{max} = A\frac{2\pi}{T}$$

and momentum is given by $p = mv$ so:

$$p_{max} = mA\frac{2\pi}{T}$$

where $m$ is the mass of the bob. And now we're done because change of mometum, or impulse, is just force times time. So if we apply a force $F$ for a (short) time $\tau$ we have $p = F\tau$ so:

$$F = \frac{mA}{\tau} \frac{2\pi}{T}$$

Just decide how long you're going to take to apply the force, and the equation above tells you what force is needed.

NB this only works if the time $\tau$ is short enough that the velocity $v$ is approximately constant while you're applying the force, and as I mentioned above this treatment only works for small oscillations. You could extend it to large oscillations by using conservation of energy to calculate $v$ at the centre of the swing - I'll leave you to have a go at that.