A hoop loop -ring) with a mass of 2 kg, radius of 60 cm is released...

Question

Question

Answer 1

Answer #1

let
m = 2 kg
R = 60 cm = 0.6 m
h = 3 m

let v is the linear speed and w is the angular speed of the hoop at the bottom of the plane.

Apply conservation of mechanical energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*m*R^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*m*(R*w)^2 = m*g*h

(1/2)*m*v^2 + (1/2)*m*v^2 = m*g*h (since v = R*w)

m*v^2 = m*g*h

v = sqrt(g*h)

= sqrt(9.8*3)

= 5.42 m/s <<<<<<<<<<-------------Answer

Answer 2

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