Respuesta :
Answer:
the speed of the mass at the bottom of its swing is 6.61m/s
Explanation:
Applying energy conservation
[tex]\frac{1}{2}m(Vlowest)^2 = mg(2R) + \frac{1}{2}m(Vtop)^2[/tex]
There is no potential energy at the bottom as the body will have a kinetic energy there.
h= 2R = 1.6m as the diameter of the circle will represent the height in the circle.
g = 9.8m/s^2
m will cancel out, so the net equation becomes.
[tex]\frac{(Vbottom)^2}{2} = 2gR + \frac{(Vtop)^2}{2}[/tex]
= [tex]2*9.8*0.8 + \frac{(3.5)^2}{2}[/tex]
= 15.68+ 6.125
[tex]\frac{(Vbottom)^2 }{2}[/tex] = 21.805
(Vb)^2 = 2*21.805
= 43.64
Vb = 6.61m/s
