Find watts from rotational kinetic energy

[Sneakatone]

a)I used the equation KE=1/2Iw^2

I=mR^2 ---> 1.5x10^30*20000^2
w=2.1*2pi
KE=1/2(1.5x10^30*20000^2)(2.1*2pi)^2=5.2x10^40 J
t=2.1 rev/s / 1.4x10^-15 rev/s^2=1.5x10^15 seconds

5.2x10^40 J/1.5*10^15 seconds =3.4x10^25 watts

B)1.15*10^15 s=1.1*10^18 years

I feel like this method is correct.

 

[Sneakatone]

edit

a)I used the equation KE=1/2Iw^2

I=mR^2 ---> 1.5x10^30*20000^2
w=2.1*2pi
KE=1/2(1.5x10^30*20000^2)(2.1*2pi)^2=5.2x10^40 J
t=2.1 rev/s / 1.4x10^-15 rev/s^2=1.5x10^15 seconds

5.2x10^40 J/1.5*10^15 seconds =3.4x10^25 watts

B)1.15*10^15 s=1.1*10^18 years

I feel like this method is correct.

 

[Dick]

a)I used the equation KE=1/2Iw^2

I=mR^2 ---> 1.5x10^30*20000^2
w=2.1*2pi
KE=1/2(1.5x10^30*20000^2)(2.1*2pi)^2=5.2x10^40 J
t=2.1 rev/s / 1.4x10^-15 rev/s^2=1.5x10^15 seconds

5.2x10^40 J/1.5*10^15 seconds =3.4x10^25 watts

B)1.15*10^15 s=1.1*10^18 years

I feel like this method is correct.

There some problems there. For one thing the moment of inertia of a sphere is (2/5)MR^2. For another you are supposed to assume the rate of energy loss is constant, not that the rate of angular deceleration is constant. Finally saying "1.15*10^15 s=1.1*10^18 years" is just plain silly. A year is much longer than a second.

 

[Sneakatone]

using the inertia equation I have 2.4x10^38.
making KE=2*10^40
new answer =1.4*10^25 watts

 

[Dick]

using the inertia equation I have 2.4x10^38.
making KE=2*10^40
new answer =1.4*10^25 watts

I don't think you are rounding correctly for the first answer if you want two significant figures. And you didn't pay any attention to my second comment. The rate of deceleration isn't a constant. The rate of energy loss is. You can't divide the energy by the expected lifetime given constant deceleration. Differentiate the KE expression to get dKE/dt in terms of dw/dt.

 

[Sneakatone]

the derivative would be Iw

 

[Dick]

the derivative would be Iw

No, that's the derivative with respect to w. You want the derivative with respect to t. Use the chain rule.

 

[Sneakatone]

I got (w^2)/2

 

[Dick]

I got (w^2)/2

How? KE=(1/2)*I*w^2. Take d/dt of both sides. I is constant. w isn't. It's a function of t.

 

[Sneakatone]

would it be just w?

 

[Dick]

would it be just w?

No!, why would you think that? You're just guessing. Use the chain rule!
dKE/dt=d((1/2)*I*w^2)/dw*dw/dt.

 

[Sneakatone]

derivative of the outside 1/2Iw^2=Iw
derivative of the inside Iw=I

so final derivative = I^2W

 

[Dick]

derivative of the outside 1/2Iw^2=Iw
derivative of the inside Iw=I

so final derivative = I^2W

The derivative with respect to w is Iw, that's the outside. You've got that. Now the 'inside' is just w. You want to differentiate that with respect to t. What is it?

You should really review the chain rule.

 

[Sneakatone]

would this involve a derivative being differentiating something with a y leaving y' on one side?
if so it would be 2Iw

 

[Dick]

would this involve a derivative being differentiating something with a y leaving y' on one side?
if so it would be 2Iw

I'll give you a closely related example. If F=Ky^3, then dF/dt=K*3*y^2*dy/dt. The K*3*y^2 is the derivative of the 'outside', the dy/dt is the derivative of the 'inside'.

Like I said you are really thrashing around on this. Review it. For now can you apply that pattern to this problem?

 

[Sneakatone]

dKE/dt=Iw*dw/dt

 

[Dick]

dKE/dt=Iw*dw/dt

Right. Once more review this. Now use that to find watts.

 

[Sneakatone]

what am I suppose to plug in for dw/dt?

 

[Dick]

what am I suppose to plug in for dw/dt?

What they gave you in the problem statement.

 

[Sneakatone]

can dw/dt also be power=F*v?
I don't know it it is given or if I should solve for it.

 

[Dick]

can dw/dt also be power=F*v?
I don't know it it is given or if I should solve for it.

dw/dt is the rate at which w is changing. If you read the problem statement, they GAVE you that. Read it again. Please?

 

[Sneakatone]

(6*10^38)*(13.19)*(1.4*10^-15)=1.1*10^25 watts

 

[Dick]

(6*10^38)*(13.19)*(1.4*10^-15)=1.1*10^25 watts

I is wrong, I'm not sure why but it looks like you just used MR^2. w is ok. Yay! dw/dt is wrong. I do know why for that one. You forgot the 2*pi. Could you please sit down and try to work this out carefully??

 

[Sneakatone]

(2.4x10^38)(13.19)(1.4*10^-15*2pi)=2.78*10^25 watts
I did not realize dw/dt had to be in rad/s^2

 

[Dick]

(2.4x10^38)(13.19)(1.4*10^-15*2pi)=2.78*10^25 watts
I did not realize dw/dt had to be in rad/s^2

Now that finally looks ok. You should pick a consistent number of significant places to round to and stick to it. But that's a minor problem for me. For getting the online checker to agree with you it might be.

 

[Sneakatone]

for the second part I guessing I am suppose to do KE/watts to get time and convert to years , is that correct?

 

[Dick]

for the second part I guessing I am suppose to do KE/watts to get time and convert to years , is that correct?

It does say energy loss rate is constant, yes? So you can do that. Now try not to bung up the conversion from seconds to years again. I think you can handle that.

 

[Sneakatone]

so I now have 0.5*(6*10^38)13.19^2/2.78*10^25 =1.8*10^15 s

 

[Dick]

so I now have 0.5*(6*10^38)13.19^2/2.78*10^25 =1.8*10^15 s

Why, why, why. Why did you go back to using 6*10^38 for I again? I know it's wrong. You should know that's wrong. We've talked about this before. I'm trying to help you understand the problem. Dealing with sloppy mistakes is not my problem. You have enough problems without doing that.