Angular speed + moment of inertia

1. Apr 7, 2013

MicahP

1. The problem statement, all variables and given/known data

Water falls onto a water wheel causing it to rotate. Consider an instant when the water wheel is initially motionless and then 100 kg of water hits tangent to the wheel at a radius of 2 m (this water can be treated like a point mass). If the moment of inertia of the water wheel is 3000 kg∙m2,what is the angular speed of the water wheel immediately after the water has hit it (in rad/s)?

Picture: (the picture gives the initial velocity as 5 m/s)

2. Relevant equations

$I=mr^2$
$KE_r=1/2Iω^2$
$Ʃ\tau = Iα$?

3. The attempt at a solution

Well, the first thing I did was find the moment of inertia of the water:

$I_w = 1/2(100kg)(2m)^2 = 200 kg*m^2$

And then I thought that I should use the kinetic rotational energy equation, but that didn't get me anywhere. I also tried to find a similar problem in my textbook, but there wasn't really anything like it.

I do know the answer is supposed to be 0.29 rad/s, if that helps.

2. Apr 7, 2013

BruceW

Hey, welcome to physicsforums :)
Ah, your answer for the moment of inertia of the water is almost correct. (Where did that 1/2 come from?) And for the next part, you don't need to use the kinetic rotational energy (who knows if that is even conserved?). But there is a quantity that is much more likely to be conserved (some would even say it is always conserved). Hint: it is not in your relevant equations, but it is definitely relevant for angular motion questions. Well, I guess it is related to the third equation in your relevant equations section.

3. Apr 7, 2013

Zatman

Hello!

Do you know about angular momentum? You should solve this using the principle of conservation of momentum.

Also, I noticed that your calculation of the moment of inertia of the water has a factor of a 1/2 in there that shouldn't be:

$I_w = m_w r_w^2 = 100\times2^2 = 400kgm^2$

4. Apr 7, 2013

MicahP

Okay, here's what I tried:

$v_t = rω$
$5 m/s = (2m)ω$
$ω = 2.5 rad/s$

$I_iω_i = I_fω_f$
$(3000)ω_i = (400)(2.5)$
$ω_i = 0.333 rad/s$

which wasn't right (but was one of the answer choices), and then I thought

$(3000)ω_i = (3400)(2.5)$
$ω_i = 2.8333 rad/s$

which...also isn't right. I feel like this is a problem that is actually really simple and I am totally missing it.

5. Apr 7, 2013

Zatman

You're actually very close! (It isn't quite the way I did it, but it works just as well).

Check the substitution of values into your last equation carefully!

6. Apr 7, 2013

BruceW

This is very close, but not quite right. (I am guessing that ω_i means the angular speed of the water wheel after the water hit it). The problem is that the 'new inertia' is not just the inertia of the water wheel. (Hint: the water doesn't just 'fall off' when it strikes the water wheel, and remember the water has mass too).

7. Apr 8, 2013

MicahP

Okay...I have spent an inordinate amount of time trying to figure this out, and I still don't follow...

8. Apr 8, 2013

Zatman

'angular momentum before' = 'angular momentum after'

You expressed this as $I_i\omega_i = I_f\omega_f$

Just think through it logically. Initially the wheel isn't turning so has zero angular momentum and contributes nothing to the LHS. You have found the M.I of the water and its angular speed before the collision. So substitute those values into the LHS.

The RHS represents the total angular momentum after the collision. What is rotating after the collision?

9. Apr 8, 2013

MicahP

Oh wow, I can't believe I didn't think of that

$I_iω_i = I_fω_f$
$(400)(2.5) = (3400)ω_f$
$ω_f = .29$

thanks, all