20. Un péndulo, que consta de una cuerda ligera de longitud L y un esfera pequeñ

Question

20. Un péndulo, que consta de una cuerda ligera de longitud L y un esfera pequeña, se balancean en el plano vertical. La cuerda golpea una clavija ubicada a un distancia d bajo el punto de suspensión. a) Demuestre que, si la esfera se libera desde una altura por abajo de la clavija, regresara a esta altura después de que la cuerda golpee la clavija. b) Demuestre que, si el péndulo se libera desde la posición horizontal (0-90°) y se balancea en un circulo completo con centro en la clavija, el valor mínimo de d debe ser 31/5 Clavija

Explanation / Answer

Let the length of the string be L and mass m.

Let the point where pendulum is released be P.

When the pendulum is at rest initially at P , it has only potential energy.

It is given by U = m g H , where H is the height or displacement of the pendulum from its equilibrium position.

It's velocity is zero as it is at rest initially at P.

When it reaches the pin , it's velocity again becomes zero as it comes to rest. Let the pin be at a point Q.

So energy possessed by the pendulum at Q is only potential energy and by using law of conservation of energy, we can say that energy at this point is equal mgH since the kinetic energy is zero. If the pin is at a height H' , potential energy of the pendulum at Q is U' = mgH'.

But we have proved that U = U' using the law of conservation of energy. So

mg H = mg H'

H = H'

So the point of release of pendulum and pin are at same height.

Since the pin is at rest and does not effect the pendulum, we can say that the pendulum reaches the same height again after hitting the pin.

B) As pin will be the center of the circle , radius will be equal to difference between the length of the pendulum and d.

So radius of the circle traced by the pendulum is

r = L - d ............. ( 1 )

When the pendulum is at the top , tension acting in the string of the pendulum is equal to its weight. So

T = mg

Since tension in the string is the only force acting in this case

T = ma

So ma = mg

a = g

where a = v2 / r is the Centripetal acceleration.

Then g = v2 / r

v2 = g r

From the equation ( 1 ) , v2 = g ( L - d )

Let the pendulum is released at a point R in horizontal position.

Energy of the pendulum at R is E = mgd ( for the minimum value of d )

At the top position , energy is E' = 1/2 m v2 + mg ( L - d )

By law of conservation of energy , E = E'

1/2 mv2 + mg ( L - d ) = mgd

1/2 m g ( L - d ) + m g ( L - d ) = mgd

( L - d ) /2 + ( L - d ) = d

d = ( 3L/2 ) - ( 3 d / 2 )

d + 3d/2 = 3 L /2

5d/2 = 3 L/2

d = 3 L /5

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