The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer: Most inhabitants of urban areas have nonagricultural jobs meaning not of, relating to, or used in farming and agriculture nonagricultural products/land. Urban areas are very developed, meaning there is a density of human structures such as houses, commercial buildings, roads, bridges, and railways.
Explanation:
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For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
