Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
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Explanation:
Given,
- m = 100 kg
- g = 10 N/kg¹
- h = 60 m
- t = 20 s
To Find:
a) Work done by the pump
b) Potential energy stored in the water
c)Power spent by the pump
d)Power rating of the pump.
Solution:
We know that,
- f = 100 kg * 10N/kg
- d = 60 m
[The unit'll be joule since N×M = J]
- b) Potential energy stored in the water
- m = 100 kg
- g = 10N/kg
- h = 60
- same condition here as well, N×M = J
- c) Power of the Pump
- where P = Power; W = Work done & T = Time taken
- As we got the value of work done on question (a),& ATQ time taken is 20 S.
- d) Power rating of the pump = 3 kW
Assumption: The pump is 100% efficient & works well.
