Respuesta :
Answer:
A)
[tex] 3.30[/tex] ms⁻¹
B)
[tex]25.05[/tex] J
C)
i)
[tex]6.60[/tex] ms⁻¹
ii)
[tex]0[/tex] ms⁻¹
iii)
[tex]4.67[/tex] ms⁻¹
Explanation:
A)
[tex]v_{cm}[/tex] = velocity of center of hoop
w = angular speed of the hoop = 3 rads⁻¹
d = diameter of the hoop = 2.20 m
radius of the hoop is given as
[tex]r =\frac{d}{2} =\frac{2.20}{2} = 1.10 m[/tex]
velocity of center of hoop is related to angular velocity by the relation given as
[tex]v_{cm} = rw[/tex]
inserting the values
[tex]v_{cm} = (1.10) (3)[/tex]
[tex]v_{cm} = 3.30[/tex] ms⁻¹
B)
[tex]I[/tex] = moment of inertia of hoop
[tex]m[/tex] = mass of the hoop = 2.30 kg
[tex]r[/tex] = radius of hoop = 1.10 m
Moment of inertia of the hoop is given as
[tex]I = m r^{2}[/tex]
[tex]I = (2.30) (1.10)^{2} = 2.783[/tex] kgm²
Total kinetic energy is given as
[tex]T = (0.5) m v_{cm}^{2} + (0.5) I w^{2}[/tex]
[tex]T = (0.5) (2.30) (3.30)^{2} + (0.5) (2.783) (3)^{2}\\T = 25.05[/tex] J
C)
i)
[tex]v_{h}[/tex] = velocity at highest point
velocity at highest point is given as
[tex]v_{h} = 2 v_{cm} [/tex]
[tex]v_{h} = 2 (3.30) \\v_{h} = 6.60[/tex] ms⁻¹
ii)
[tex]v_{b}[/tex] = velocity at bottom
velocity at bottom point is given as
[tex]v_{b} = v_{cm} - v_{cm} [/tex]
[tex]v_{h} = 3.30 - 3.30 \\v_{h} = 0[/tex] ms⁻¹
iii)
[tex]v_{r}[/tex] = velocity at right
velocity at right is given as
[tex]v_{r} = \sqrt{2} v_{cm} \\v_{r} = \sqrt{2} (3.30)\\v_{r} = 4.67[/tex] ms⁻¹
