Circular/rotational motion problem - Waterwheel with bucket

[Oshada]

Homework Statement

A water-wheel equipped with a single bucket is fixed in the orientation shown in the diagram below, while the bucket is filled with water. When the bucket is full, the wheel is released so that it can rotate freely. The wheel has a uniform density and a mass of 100 kg, radius 10m and the bucket has a capacity of 20 litres of water (with mass 20 kg). The mass of the bucket is negligible compared to the mass of the water.

(a) Determine the initial angular acceleration of the water wheel immediately after it is
released (i.e. when θ = 0).

(b) Determine the torque experienced by the wheel as a function of 9 over the range from
θ = 0 to θ = π/2 assuming no water is lost from the bucket.

(c) Determine the final angular velocity of the wheel assuming the water runs out of the
wheel at θ = π/2.

(d) Assume the wheel has four identical buckets to the one in the diagram, attached to the
rim with equal spacing. The buckets are all filled and emptied on the same basis as
described above. If the wheel is attached to an electric generator that exerts a torque of
1000 Nm on the wheel, and the wheel then rotates at an average angular speed of
0.5 rad/s, determine the average power being generated by the wheel.

Homework Equations

All the usually relevant circular motion equations involving θ, I, ⍺, τ and ω

The Attempt at a Solution

For part a), I found τ from τ = r x F and found ⍺ from τ = I⍺. Since v = 0 at the start I ignored centripetal force. I got ⍺ = 0.40 rad/s^2, but the answer given is 0.39 rad/s^2. Due to this I haven't really worked out the rest of the sections. Also, I'm lacking a plan of attack for section c). In section d), since our course did not include P = τ.ω I think I might have to derive it.

Any help is most appreciated!

 

[peterpang1994]

a)Let the moment of inertia of the wheel = m_wr²/2

The moment of inertia of the bucket = m_br²

The torque acting on the system = 20gr=2000

Using the equation,
Torque = moment of inertia X angular acceleration
200r = (20r+100/2r² ) X angular acceleration
angular acceleration = 0.38rads^-2

 

[Oshada]

Thanks a lot! However, the answer was 0.39 rad/s^2, which makes me think that the moment of inertia of the bucket might be something different?

 

[peterpang1994]

But I think that the bucket and the water inside can be consider as point mass, the little difference can be ignored, right?

 

[peterpang1994]

In section c), you have to use the result from b) and using integration by in ⍺dt = ω+ω₀

 

[Oshada]

Thanks a lot! If you can show me how to get torque in terms of angular displacement it'd be very helpful :D

 

[peterpang1994]

b) τ = R X F R is the perpendicular distance between the force vector and the centre of rotation
τ = mgrcosθ
=2000cosθ

where 0<= θ<= π/2

 

[Oshada]

OH right, I didn't think about the r changing D: Thanks a lot! (The answer is 1960cosθ but that's just g = 9.8 m/s^2). As for section c), how can I use the factor that all the water has run out at the bottom (θ = π/2)?

 

[peterpang1994]

In c) the assumption of all water run out at bottom means that there is no water run between θ=0 and θ=pi/2
Do you want me to show you how to calculate c)?

 

[Oshada]

The answer to c) is 0.87 rad/s. I think the water runs out gradually, from 0 to π/2.

 

[peterpang1994]

That may not be the case, the water should run out when θ=pi/2, I tried to calculate(base on my assumption), I get the right answer

 

[Oshada]

Which formulas did you use? If you could show your working that'd be great!

 

[peterpang1994]

I have used integration for c)

⍺=τ/I = 1960cosθ / (20r+100/2r2 ) = (49/130)cosθ

ω= ∫ ⍺ dt
= ∫ (49/130)cosθ dt
= 49/130 ∫ cosθ dt
Because ω=dθ/dt

ω= 49/130 ∫ (cosθ)/ω dθ

derivative both side,

dω/dθ = 49/130 (cosθ)/ω
ωdω = 49/130 cosθdθ

integrate both side

∫ ωdω = 49/130 ∫ cosθdθ
(1/2)ω² = (49/130)sinθ
ω = √(49/65)sinθ
when θ=π/2,
ω = √(49/65)sin(π/2) = 0.87rads^-1

 

[Oshada]

Oh wow, that's tricky... is that the only way to do it? But thank you very much!