A holiday ornament in the shape of a hollow sphere with mass 0.015 kg and radius 0.055 m is hung from a tree limb by a small loo
1 answer:
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Answer:
a)1.93 kg-m^2
b) 1.45 kg-m^2
c) = 0
d) 1.15 kg-m^2
Explanation:
mass of the bar M = 4 kg
length of the bar = 2 m
mass of balls m1= m2= 0.3 kg
moment of inertia of bar
about an axis perpendicular to the bar through its center.
a) MOI of bar + 2×m×(L/2)^2
+
now putting the values of m, M and L as above and solving we get
I= 1.93 kg-m^2
b) perpendicular to the bar through one of the balls
= 1.45 kg-m^2
c) parallel to the bar through both balls
zero as the no mass distribution along the parallel to the bar through both balls.
d) parallel to the bar and 0.500 m from it.
I=
putting values and solving we get
1.15 kg-m^2
The time elapsed when the ball reaches the window is 1.55s
As per the question:
Final velocity, v = 11 m/s
Height, h = 29m
Acceleration, a = g = -9.8 m/s
From the third equation of motion:
v² = u² +2as
u² = v² - 2as
u² = (11)² - 2 × (-9.8) × (29)
u² = 689.4
u = √689.4
u = 26.25 m/s
Now, from the first equation of motion:
v = u + at
t = 1.55 s
Therefore, the time elapsed when the ball reaches the window is 1.55s.
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