If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be
1 answer:
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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:
part (a)
Part (b)
Explanation:
Given,
- mass of the smaller disk =
- Radius of the smaller disk =
- mass of the larger disk =
- Radius of the larger disk =
- mass of the hanging block = m = 1.60 kg
Let I be the moment of inertia of the both disk after the welding,
part (a)
A block of mass m is hanging on the smaller disk,
From the f.b.d. of the block,
Let 'a' be the acceleration of the block and 'T' be the tension in the string.
Net torque on the smaller disk,
From eqn (1) and (2), we get,
part (b)
In this case the mass is rapped on the larger disk,
From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,
Let '' be the acceleration of the block in the second case,
From the above expression,
