Question

A cylindrical rod 21.5 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 6.90 cm and a mass of 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod.
(a) After it falls through 90°, what is itsrotational kinetic energy?
J
(b) What is the angular speed of the rod and ball?
rad/s
(c) What is the linear speed of the ball?
m/s
(d) How does this compare with the speed if the ball had fallenfreely through the same distance of 24.8 cm?
vswing is ---Select---greater thanless thanvfall by %

Answer 1

Answer

given,

length of rod = 21.5 cm = 0.215 m

mass of rod (m) = 1.2 Kg

radius, r  = 1.50

mass of ball, M = 2 Kg

radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m

considering the rod is thin

$I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]$

$I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]$

     I = 0.144 kg.m²

rotational kinetic energy of the rod is equal to

$KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2$

$KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2$

  KE = 6.15 J

b) using conservation of energy

   $K_f + U_f = K_i + U_i + \Delta E$

   $\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0$

   $\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15$

    ω = 9.25 rad/s

c) linear speed of the ball

     v  =  r ω

     v  =  (L+R )ω

     v  =  (0.215+0.0345) x 9.25

     v =2.31 m/s

d) using equation of motion

  v² = u² + 2 g h

  v² = 0 + 2 x 9.8 x 0.248

   v = √4.86

  v =2.20 m/s

speed attained by the swing is more than free fall

  % greater = $\dfrac{2.31-2.20}{2.20}\times 100$

                   = 5 %

speed of swing is 5 % more than free fall