Engineering
Physics
SHM of Simple Pendulum
Collision
Acceleration
Question

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position and released. The bob hits the wall with restitution coefficient being 25. What is the minimum number of impacts with the wall after which the maximum angular displacement becomes less than 60°?

 

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Solution
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When the bob is at athe horizontal position A, its height above B = length of the string =  ℓ . Therefore, potential energy at A = mgℓ . When the bob is released, it hits thewall at B and the enitre potentialenergy is converted into kinetic energy. If v is the velocity with which the bob hits wall, then

12mv2=mgℓ  or  v=2gℓ                                   ... (i)

Speed of the bob after the first rebound,

v1 = ev

Speed of the bob afte the second rebound,

v2 = ev1 = e2v               

Speed of the bob after n rebounds, vn = env

If the ball rises to a position C at a height

h = BD afte n rebounds, then from the principle of conservation of energy, we have

12mvn2=mgh

h=vn22g=(env)22g = e2nv22g   ... (ii)

Using eqn. (i) in (ii), we get

h=2gℓe2n2g=ℓe2n                                               ... (iii)

If θn is the angle the string subtends wtih vertical after n rebounds,it follows from figure that

h = OB – OD = ℓ – ℓ cos θn = (1 – cos θn) ...  (iv)

From eq. (iii) and (iv), we have

ℓ e2n = ℓ (1 – cos θn) or e2n = (1 – cos θn)

From qn to be less than 60°, i.e. cos qn is greate than 12,

  (1 – cos θn) must be less than 12  

  Thus, e2n < 12

  Given,e=25

  =(25)2n< 12   or (45)n<12

or (0.8)n < 12 or  n ln  (0.8) <  ln (0.5)

or – (0.0969) n < – 0.3010 or n > 0.30100.0969

or         n > 3.1

The least integer treater than 3.1 is 4. Hence n = 4. Thus, after 4 impacts,the amplitude will becomes less than 60°.