Please Help!

[the_sunshine]

A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 575 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 4500 rev/min.

 

(a) Find the kinetic energy stored in the flywheel.

 

(b) If the flywheel is to supply energy to the car as would a 10.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.

 

I tried:

 

Part a)

w=4500rev/min*2PR/60= 706.5m/sec

 

I=mass*R^2=575kg*(1.50m)^2= 1293.75 kg*m^2

 

Then, KE=1/2*I*w^2 = 1/2 (1293.75 kg*m^2) * (706.5 m/s)^2 =

322 895 153.9 J

 

Part b)

Power = d*w/d*t

 

And it is wrong. I cannot go to part b with incorrect answer in a. Can anyone please try to explain to me what am i doing wrong?

[thedarkshade]

http://www.youtube.com/watch?v=Vn32inxsChc try this!

[the_sunshine]

so is it my mistake in the formula? KE=1/2 m*v^2

i'm sorry, i still cannot understand.

could you be more specific?

[thedarkshade]

so is it my mistake in the formula? KE=1/2 m*v^2

i'm sorry, i still cannot understand.

could you be more specific?

The formula for linear kinetic enery is:[math]KE=\frac{mv^2}{2}[/math] and for rotational kinetic energy just instead of m put momentum and instead of v put omega (angular velocity)

[the_sunshine]

yes, but it's still wrong

[thedarkshade]

the video gives you a good explanation related to energy,, so it might help on that part!

 

yes, but it's still wrong

WHAT IS WRONG??

[the_sunshine]

the answer for KE i got

[thedarkshade]

the answer for KE i got

Oh , ok then,, the formula I posted above is the correct one, if you've watched the video, it clearly tells you even the derivation.

[the_sunshine]

ok, thanks!

[thedarkshade]

ok, thanks!

Glad to be useful!

[swansont]

I=mass*R^2=575kg*(1.50m)^2= 1293.75 kg*m^2

 

mr^2 is the moment of inertia for a wheel with all the mass at the rim. Is that what the flywheel looks like? If it's a solid disc, the equation is 1/2 mr^2

[Mr Skeptic]

[math]KE=\sum \frac{1}{2}mv^2 = \sum \frac{1}{2}m(r^2 \omega^2) = \frac{1}{2} (\sum mr^2) \omega^2 = \frac{1}{2}I \omega^2[/math]

 

The rotational kinetic energy is really the sum of the kinetic energies of its constituents (from the center of mass reference frame).

 

The moment of inertia depends on the mass distribution. For a ring (or a hollow cylinder), all the mass would be concentrated at R, so the moment of inertia would be [math]I = \sum mr^2 = MR^2[/math].

 

However, a flywheel is usually a disk (or filled in cylinder). Since some of the mass is located at a radius r < R, a disk will have a smaller moment of inertia than a ring of the same radius. The moment of inertia would be like that of a set of concentric rings. If the disk was of constant density, each ring would have mass [math]m \approx \frac{M}{\pi R^2} 2\pi r dr[/math] proportional to the circumference. Overall, it would be [math]I = \int_0^R mr^2 dr = \int_0^R \frac{M}{\pi R^2} 2\pi r^3 dr = \frac{2M}{R^2} \int_0^R r^3 dr = \frac{2M}{R^2} \frac{1}{4}(R^4 - 0) = \frac{1}{2}MR^2[/math], half of what it would have been for the ring.

 

You'd probably want to ignore my dusting up of my rotational mechanics, and just look the thing up on a table. There are tables for all the simple shapes, and you wouldn't want to calculate I for yourself (except to understand what is going on, or as practice).

 

Unless your flywheel is a ring, your mistake was the value of I.

[thedarkshade]

Part a)

w=4500rev/min*2PR/60= 706.5m/sec

So you have 4500 rev/min, that's the frequency.You just need to multiply [math]2\pi[/math] with 4500rev/60s. then you'll get [math]\omega=2\pi\frac{4500rot}{60s}[/math]

[math]\omega=2\times 3.141592654 \times 75[/math]

and you get: [math]\omega=471.2388981\frac{rad}{s}[/math], that's the angular velocity!

 

to turn them in linear velocity you use: [math]v=\omega\times r[/math] and then just replace the datas:

[math]v=\omega\times r[/math]

[math]v=471.2388981 \times 1.50=706.85834715\frac{m}{s}[/math] which means that part you did is right!

 

to find [math]I[/math] you have [math]EK=\frac{1}{2}mr^2[/math] (like swansont said) then you have: [math]I=\frac{1}{2}mr^2[/math]

[math]I=\frac{575kg\times (1.50m)^2}{2}[/math]

[math]I=\frac{575kg\times 2.25m^2}{2}[/math]

[math]I=\frac{19293.75kgm^2}{2}[/math]

[math]I=646.875kgm^2[/math]

 

and about kinetic energy [math]EK=\frac{1}{2}mv^2[/math]

then you get: [math]EK=\frac{1}{2}mv^2[/math]

[math]EK=\frac{575kg \times 706.85834715 (m/s)^2}{2}[/math]

[math]EK=\frac{499648.7228J}{2}[/math]

[math]EK=249824.3614L[/math]