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In the given figure $L = 1.25 \mathrm m$ and $\theta = 40^\circ$ What is the least value that $v_\circ$ can have for the bob of the pendulum to swing down and then swing up to a) a horizontal position, b) a vertical position with cord remaining straight ?
enter image description here

(Sorry for this monster pic I don't intend to do this but I have no clue as to how to use Inkscape.)

My understanding of the question tells me that i have to find the minimum $v_0$ for the pendulum to reach the red positions.


For part a) I did the following,

Reference point for the potential energy : - Lowest position the pendulum.

System :- Earth and pendulum.

Initial height of the pendulum from the reference point = $L - L\cos 40 \approx 0.3$

Answer:- $$\begin{align}\\ 0 &= \triangle K + \triangle U \\&=\frac 12 m(0 - v_0^2) + mg(1.25 - 0.3)\end{align}$$
$$\implies\frac 12 v_0^2 = g \times 0.95 \implies v_0 = 4.31 \mathrm {m/s}$$

For part b) I did the following :-

Reference point for the potential energy : - Lowest position the pendulum.

System :- Earth and pendulum.

Answer:- $$\begin{aligned} 0 &= \triangle K + \triangle U \\ &=\frac 12 m(0 - v_0^2) + mg(2.5 - 0.3)\end{aligned}$$
$$\implies\frac 12 v_0^2 = g \times 2.2 \implies v_0 = 6.5 \mathrm {m/s}$$


My answer to (b) is incorrect :( .

Calculation given in the book :-

Let $u = \sqrt{Lg}$

$$\begin{aligned} 0 &= \triangle K + \triangle U \\ &=\color{red}{\frac 12 mu^2} - \frac 12 mv_0^2 + mg(2.5 - 0.3)\end{aligned}$$
$$\implies\frac 12 v_0^2 = g \times 2.2 + {g\times 1.25 \over 2} \implies v_0 = 7.45 \mathrm {m/s}$$


My confusions:

  1. I know $u$ is the final velocity of the pendulum and the highest point but does not know why we considered it and how did we get its value?

  2. Should not the velocity of the bob at the highest point is 0 and the KE 0 for minimum value of $v_0$?

  3. Why did not we get a similar term for question a)?

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  • $\begingroup$ @Qmechanic Is classical mechanics wrong tag to use ? $\endgroup$ – A---B Dec 28 '16 at 21:15
  • $\begingroup$ The classical mechanics tag should not be applied if the newtonian mechanics tag suffice. $\endgroup$ – Qmechanic Dec 28 '16 at 21:19
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The issue in your solution is that you didn't considered that the cord has to remain straight.

If you think at the forces acting on the mass at the highest point, you can say that the centripetal force should be equal to the gravitational one, in order for the mass not to have any force acting on it; that is the minimum energy condition for which the cord is straight. $$\frac{m u^2}{L} = m g \ \Rightarrow\ u =\sqrt{gL} $$ This leads you to the correct result.

This isn't right for point (a) because there there isn't the request about the rope to be straight.

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  • $\begingroup$ So you considered the tension to be zero, right ? Also if in the first part it was asked that the cord should be straight then $T = {mv^2\over L}$ as $T = 0$ therefore $v = 0$. Am I correct ? $\endgroup$ – A---B Dec 28 '16 at 21:25
  • $\begingroup$ @A---B yes for both questions. In the first case, because what is required is the minimum velocity to get that condition, in the second because tension and gravitational force act on different directions. $\endgroup$ – JackI Dec 28 '16 at 21:28
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    $\begingroup$ Thank you very much. I was struggling so much with this question. Puff, thanks. you are a genius. Happy new year to you $\displaystyle \huge \ddot \smile$. $\endgroup$ – A---B Dec 28 '16 at 21:30

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