Answer:
u = 3.35 m/s
Explanation:
given,
mass , m = 0.455 kg
R = 0.675 m
Height of Loop = 1.021 m
the speed required at the top of loop be v
equating the force vertically
[tex]m g =\dfrac{mv^2}{r}[/tex]
[tex]9.81 =\dfrac{v^2}{0.675}[/tex]
v² = 6.622
v = 2.57 m/s
Let the initial speed of ball be u
using conservation of energy
[tex]\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)[/tex]
where, [tex]I =\dfrac{2}{5}mr^2[/tex]
[tex]\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)[/tex]
[tex]0.7 u^2 + g H = 0.7 v^2 + g(2R)[/tex]
[tex]0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)[/tex]
0.7 u² = 7.85092
u² = 11.2156
u = 3.35 m/s
the initial speed is 3.35 m/s
