The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Hard dynamics question!

Hi I'm pretty stuck on the following, if anyone could help that would be great:

A rotating disc of mass M, radius R and angular velocity ω has its mass distributed uniformly over its area. Calculate the total kinetic energy as follows:

i) Write down the mass of the disc contained between radius r and radius r + dr.
ii) Write down the speed of this mass.
iii) Calculate the kinetic energy of this mass.
iv) calculate the total kinetic energy of the whole disc by integrating the previous with respect to r between the limits r = 0 and r = R.

I think I've done the first two parts correctly but am seriously struggling with part iv. I have been given the answer to one of the parts(iii or iv I presume): (1/4)MR^2ω^2

Thanks in advance
Original post by iamaboy418
Hi I'm pretty stuck on the following, if anyone could help that would be great:

A rotating disc of mass M, radius R and angular velocity ω has its mass distributed uniformly over its area. Calculate the total kinetic energy as follows:

i) Write down the mass of the disc contained between radius r and radius r + dr.
ii) Write down the speed of this mass.
iii) Calculate the kinetic energy of this mass.
iv) calculate the total kinetic energy of the whole disc by integrating the previous with respect to r between the limits r = 0 and r = R.

I think I've done the first two parts correctly but am seriously struggling with part iv. I have been given the answer to one of the parts(iii or iv I presume): (1/4)MR^2ω^2

Thanks in advance


Hello and welcome to TSR physics.

Maybe if you could write your answers to parts i and ii it would be clearer.
We can first check if these are correct.

For part 3 you just do ½mv² on the mass you found in part2 and remember that v = rω

Do you know, then, how to integrate?

The answer you have been given is the final answer for the ke of rotation of the disc.
(edited 10 years ago)
Reply 2
Avatar for iamaboy418
iamaboy418
OP
2
Thank you for replying!
Well for part 1 I used mass=density*area, so I got ro*pi*((r+dr)^2-r^2)
for part ii I used v=r*omega which game me (r+dr)* omega.
And then for kinetic energy I got 1/2*ro*pi*((r+dr)^2-r^2)*((r+dr)*omega)^2
I don't understand how that integral then becomes (1/4)MR^2ω^2
Original post by iamaboy418
Thank you for replying!
Well for part 1 I used mass=density*area, so I got ro*pi*((r+dr)^2-r^2)
for part ii I used v=r*omega which game me (r+dr)* omega.
And then for kinetic energy I got 1/2*ro*pi*((r+dr)^2-r^2)*((r+dr)*omega)^2
I don't understand how that integral then becomes (1/4)MR^2ω^2


You need to consider the strip (ring) of disc as being of width dr and length equal to the circumference 2πr
This strip has an area equal to its length, 2πr times its width dr
Its mass M is therefore m.2πr dr
where m is the mass per unit area.
The kinetic energy of this strip is
½Mv²
where M is the mass in the expression above
Sub for M and for v=rw

You then integrate this expression wrt r
finally use the fact that the mass of the whole disk is its area times m
(edited 10 years ago)
Reply 4
Avatar for iamaboy418
iamaboy418
OP
2
but I still don't get the correct answer of (1/4)MR^2ω^2 after that. I get (1/4)piMR^4w^2. Do I then divide by pi*R^2?
Original post by iamaboy418
but I still don't get the correct answer of (1/4)MR^2ω^2 after that. I get (1/4)piMR^4w^2. Do I then divide by pi*R^2?


You should get after integration, for the kinetic energy E of the whole disk

E=m2πR4ω24E = \frac{m2\pi R^4\omega^2}{4}

but you know the mass of the whole disk, Md=πR2mM_d= \pi R^2 m

which means the kinetic energy is

MdR2ω24 \frac{M_d R^2 \omega^2}{4}

There is some confusion, partly my fault, because I used M for the mass of the strip.
If so you need a different M for the mass of the whole disk. Here I've used Md

BTW
It will help a lot if you learn to use "Latex" to make your formulas more readable.
People are more likely to reply if you do.
(edited 10 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you