Question

17. 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:

We are given that:

Length of string = 20 cm = 0.2 m

Mass of bob = 100 g

To find K.E and P.E

Solution:

T = 2π √ l cosθ / g

cos θ = T^2 g^2 / 4 π^2 l

Cos θ  = (0.8)^2 x (π^2) / 4 π^2 x 0.2

Cos θ = 0.8

P.E = mgl ( 1 - cos θ)

K.E = 1/2 mv^2

K.E = P.E

V^2 = 2gl ( 1 - cosθ)

V =  √ 2gl ( 1 - cosθ)

V = √  2 x 9.8 x 0.2 ( 1 - 0.8) = 0.8876 m/s

K.E = 1.2 mv^2 = 1/2  0.1 x 0.8876 = 0.0443 J

P.E =  mgl ( 1 - cos θ)

P.E = 0.1 x 9.8 x 0.2 ( 1 - 0.8)

P.E = 0.0394 J

Answer 2

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