From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
m is mass in kg,
g is acceleration due to gravity (9.8 m/s²)
h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters high.
The bouyancy force is:
Since the wood-lead system is completely submerged, the bouyancy force
is FB = ĎwgVl + ĎwgVb, where Ďw is the density of water,Vl
is the volume of
the piece of lead and Vb is the volume of the wooden block. The weight of the
combined lead and wooden block is: W = ĎlgVl + ĎbgVb. Since the system is
in equilibrium, the bouyancy force must be equal to the total weight:
ĎwgVl + ĎwgVb = ĎlgVl + ĎbgVb
now we can solve for the volume of lead:
ĎwgVl â’ ĎlgVl = ĎbgVb â’ ĎwgVb
Vl(Ďw â’ Ďl) = Vb(Ďb â’ Ďw)
Vl =
Ďbâ’Ďw
Ďwâ’Ďl
Vb
Now we substitute the values for the density of lead Ďl = 11.3 Ă— 103kg/m3 ,
the density of the wood and the density of water Ďw = 1000kg/m3
. We get:
Vl =
600â’1000
1000â’11300
(0.6m Ă— 0.25m Ă— 0.08m) = 4.66 Ă— 10â’4m3
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
