A water-wheel equipped with a single bucket is fixed in the orientation shown in

Question

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 10 m 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. Determine the final angular velocity of the wheel assuming the water runs out of the wheel at theta- pi/2.

Explanation / Answer

This is a non uniform acceleration problem. We first need to find the force/torque induced by the bucket of water. F = m*g * cos(Theta) Next we find torque: T = F x r T = 20*9.8*10*cos(theta) T = 1960 * cos(Theta) We need to relate the torque to acceleartion: T = I * Alpha Alpha = T/I Next find the moment of inertia: See wiki for a list of moments of inertia I = m*r^2 / 2 I = 1000*10^2 / 2 I = 5000 Now relate the acceleartion to position from previous equations: Alpha = 1960/5000 * cos(theta) Alpha = .392 cos(Theta) We now need to use calculus to find the final angular velocity: From here on out w will be angular velcoity (omega), a will be angular acceleratoin (alpha) and 0 will be angular position (theata). a(0) = dw/dt --> alpha as a function of thetga = d(omega)/dt use chain rule: dw/dt = dw/d0 * d0/dt since d0/dt is equal to w (omega) and dw/dt is equal to a (alpha) a(0) = w dw/d0 separate: a(0) d0 = w dw int(w dw) = int( a(0) d0 ) w^2 / 2 = int(.392cos(theta) d0) from 0 --> pi/2 w^2 / 2 = .392 sin(pi/2) w^2 = .784 w = .885 rad/sec If you need more explaination just let me know.

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