Respuesta :
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
[tex]I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2][/tex]
[tex]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][/tex]
I = 0.144 kg.m²
rotational kinetic energy of the rod is equal to
[tex]KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2[/tex]
[tex]KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2[/tex]
KE = 6.15 J
b) using conservation of energy
[tex]K_f + U_f = K_i + U_i + \Delta E[/tex]
[tex]\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0 [/tex]
[tex]\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15[/tex]
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 = [tex]\dfrac{2.31-2.20}{2.20}\times 100[/tex]
= 5 %
speed of swing is 5 % more than free fall
