Question

A pendulum consisting of a massless
string of length 20 cm and a tiny bob
of mass 100 g is set up as a conical
pendulum. Its bob now performs 75 rpm.
Calculate kinetic energy and increase in
the gravitational potential energy of the
bob.

Answer 1

Answer:

K.E   = 0.45 J    

∆P.E = 0.04 J  

Explanation:

Given data :-

 string of length L = 20 cm = 0.2 m

 tiny bob of mass m = 100 g = 0.1kg

Frequency of revolutions  n = 75 rpm = 75/60 = 1.25 rps

What we have to find out:-

1)kinetic energy K.E = ?

2)increase in the gravitational potential energy of the bob ∆P.E =?  

Since the frequency of revolution n is 1.25 rps hence time period of pendulum is T = 1/n =1/ 1.25 = 0.8 sec, then  

$T = 2 \pi \sqrt{L Cos (theta) /g}$

cos θ = ( T2 g/ 4 π 2 L)

 = ( 0.82 *10 / 4 *10* 0.2 )

 = 0.8

 θ = cos -1 (0.8 )

    = 36.86 degree

kinetic energy of bob is given by  K.E  = ½( mv2 )  

                   = ½(mr2 w2 )

     = ½( m(Lsin θ)2 (2 π/ T  )2

K.E = ½( 0.1 (0.2 sin 36.86 0 )2 (2 (√10)/0.8)2

   = 0.449 J

 K.E   = 0.45 J    

2) h = L – L cos Ө = L (1 – cos Ө )

∆P.E = mgh = mg L (1 – cos Ө ) = 0.1 * 10 * 0.2  (1 – cos θ ) = 0.2* (1 – 0.8)

  = 0.2 * 0.2 = 0.04 J  

∆P.E = 0.04 J