Vertical Circular Motion

Question

Please help, I'm struggling to find how to proceed with Energy methods, as I don't know how to get from the impulse given to a suitable initial velocity to use.

"A particle, A, of mass 500g is attached to one end of a light inextensible string of length 1 m. The other end of the string is attached to a fixed point B, vertically above A.

A is hit with a blow of impulse 1.5 N in a horizontal direction. Find the angle that the string makes when the particle first comes to rest." $$$$ This is my thought process:

Energy before is equal to Energy after. GPE at the start can be treated as 0 and KE at the start is equal to $\ 0.5mu^2$ and therefore $\ 0.25u^2$.

GPE at the end must be $ { 0.5g(1-cos\theta)}$ and KE at the end must be zero as the particle is at rest.

Equating these gives me: $ cos\theta = 1 - u^2/2g$

I don't know, however, how to get $u$, as I'm only given impulse. From that I can work out $a$ but then get stuck finding $u$. Have I gone wrong?

Note - The answer given at the back of the book is $ \theta = 56.6 $

Thanks

You could try linear impulse momentum theorem in horizontal direction — Rohan Shinde, Feb 16 '18 at 02:04
Refer to https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec09.pdf for information about linear impulse momentum theorem. — Rohan Shinde, Feb 16 '18 at 02:49
Small point of rigor. An impulse does not have units of Newtons (N), that would correspond to a force. Do you mean an impulse of 1.5 N$\cdot$s? — Anders Beta, Feb 16 '18 at 02:51
Well, the book reads "a blow of impulse 1.5 N" so I'm not quite certain :/ — Antonios, Feb 16 '18 at 02:58
and(p)/dt = F , mp_f = Integral F dt. The force is applied during what time ? — Peter Szilas, Feb 16 '18 at 10:49

Szeto · Answer 1 · 2018-02-17T02:59:11.603

When given $\theta =56.6 degrees$, I did a sneaky backward calculation and found that $u$ is approximately $3 m/s$.

This agrees with the following idea:

Initially, $1.5 kg m/s$ of momentum is given to the particle A.

By the conservation of momentum: $$p_{initial}=p_{final}$$$$1.5 =mu=0.5u$$

Thus $u=3 m/s$.

In addition, I think the author means $1.5 Ns$ instead of $1.5N$. That’s a typo.

————————

Today, I realized that there might be a more satisfying answer.

If you take the words in the question literally, that might be not a typo.

“A blow of impulse of 1.5N.”

What does that really mean?

‘Blow’, in mathematics, often mean infinity.

But how can you have 1.5N while having infinity?

The way out is Dirac delta function.

Consider a force-time graph. At t=0, there is a very strong force which lasts for a very short time that gives momentum to the particle.

The force must not be finite. If the force is finite, recall the equation $$p=F\Delta t$$ If the force is finite, then $$\lim_{\Delta t\to 0}F \Delta t = 0$$ then $p=0$ as well, then the particle would not be moving at all.

So we have concluded the force is infinite. However, how could we express the ‘1.5N’?

In layman’s term, here is the explanation. Although the force is infinitely large, we can discuss the magnitude of the infinite-ness of the force, and in this case, the magnitude of infinite-ness is 1.5N. This idea is abstract and hardly convincing, but this is the easiest way to get used to it.

Therefore, we will say the force-time relationship can be described by $$F(t)=1.5\delta (t)$$ where $\delta (t)$ is the Dirac delta function.

Dirac delta function equals infinity at $t=0$ and equals zero for other $t$.

Now, we are interested in the total momentum. By definition: $$p_{total}=\int^\infty_0 F(t)dt =\int^\infty_0 1.5\delta (t)dt$$

Note that by definition $$\int^\infty_0 \delta (t)dt=1$$

Therefore, $p_{total}=1.5Ns$, just as expected!

Vertical Circular Motion

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