Answer:
v = 3.6 m/s
Explanation:
given,
Inclination of the ramp, θ = 33⁰
Distance of the rope, d = 1.70 m
initial speed = ?
using conservation of energy
[tex]\dfrac{1}{2}mv^2 + \dfrac{1}{2}I_{com}\omega^2 = m g h[/tex]
h = d sinθ
h = 1.7 x sin 33° = 0.926 m
moment of inertia of the solid ball
[tex]I_{com}= \dfrac{2}{5}MR^2[/tex]
we know, v = r ω
[tex]\dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{2}{5}mR^2\times \dfrac{v^2}{R^2}= m g h[/tex]
[tex] \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 = mgh[/tex]
[tex]gh= \dfrac{7}{10}v^2[/tex]
[tex]v = \sqrt{\dfrac{10}{7}gh}[/tex]
[tex]v = \sqrt{\dfrac{10}{7}\times 9.8\times 0.926}[/tex]
v = 3.6 m/s
Hence, initial speed of the solid ball = 3.6 m/s
