# Lab finding the mass of a flywheel

1. Jul 17, 2014

### onelove8187

1. The problem statement, all variables and given/known data
The problem is I need to estimate the mass of a (solid equally distributed mass) disc which has an axle on which a string is tied and then attached to a (hanging) 1.022 kg mass. We wound the wheel up moving the hanging mass to a height of 0.64 m off the ground. At this point we recorded the time it would take for the mass to fall to the floor (spinning the wheel in the process). for this first run we have an average time t=10.42s. We did this several times varying the height off the floor of the hanging mass. We also ran several tests varying the mass. I have spent probably ten hours banging my head against a wall, I do not know what mistake I am making. This is what I tried to solve for the mass of the wheel in the first run:

E_pot=E_k+E_rot

E_pot=mgy
E_k(at bottom)=.5mv^2
E_rot=.5Iω^2 where I=.5MR^2 and ω=v/R

where m=known mass and M=mass of wheel

Combining all this I got:

mgy=.5mv^2+.5(.5MR^2)(v/R)^2

simplifying to

M=4m(gy-.5v^2)/v^2
for the first fun it gives≈1600kg as the answer which can't be right!

I also tried to solve this from a torque perspective and got the same answer!
possible other relevant equations:

angular acceleration=torque_net/Moment of Inertia
a_y=angular acceleration
Torque=Tension*d (where d=R in this case)

More than anything I want to see where my error in logic is here so I can better understand the material. Any help will be greatly appreciated. Thank you!

2. Jul 17, 2014

### SammyS

Staff Emeritus
Hi onlove8187. Welcome again to Physics Forums.
By the way, how did things work out with the first problem you posted? I noticed you did not respond to any of the replies given to your post.​

For this problem you found the following expression for the mass of the disc.

M=4m(gy-.5v2)/v2

This may well be the correct result.

You didn't mention how you determined v. I assume that's some velocity. Exactly what is that velocity?

Last edited: Jul 17, 2014
3. Jul 17, 2014

### onelove8187

Thanks very much for responding. I found the velocity using kinematics. We recorded the time it took the block to fall to the ground and then with the known height and an initial velocity of 0 I found the acceleration using v_f=v_i+a(Δt). After that I found the final velocity with (V_f)^2=(V_i)^2+2a_y(Δx). My thinking is that at this point the final Kinetic Energy + rotational Kinetic energy should have come completely from the gravitational potential energy that was present when the block was at height h.

4. Jul 17, 2014

### SammyS

Staff Emeritus
What did you use for the acceleration ?

In the equation, vf = vi + a(Δt), which you used to find a, you don't know vf either, do you ?

5. Jul 17, 2014

### onelove8187

You are absolutely correct. My mistake, I used y_f=y_i+v_i *t+.5a_yt^2 where y_f is 0 and v_i is 0. Then I used that acceleration to find final velocity. I've been thinking, will the string being attached to the an axle change the approach. I think that may be where my mistake is. I sort of modeled it as if the string were wrapped around the disk and not around an axle in the center.

6. Jul 17, 2014

### SammyS

Staff Emeritus

Was the string was wrapped around the axle, or was it wrapped around the disc? In the former case you need to know the radius of the axle. Also, the analysis is more involved if the string wraps over itself so many times so that the effective radius "seen" by the string decreases significantly as the string unwinds.

Is the string longer than the initial distance of the hanging mass from the floor?

By the way:

If you can assume constant acceleration, then the average velocity is given by vavg = (vf + vi)/2

7. Jul 17, 2014

### onelove8187

The string was wrapped around the axle and we do know the radius of the axle. Also, it didn't wrap over itself too much changing the radius felt by the string. Lastly, yes the string was longer than required to reach the floor.

8. Jul 17, 2014

### SammyS

Staff Emeritus
So there's some additional algebra to do.

If the radius of the axle is r, then that must replace R in at least one of your basic equations.

Show what you get now.

9. Jul 17, 2014

### onelove8187

Ok I replaced the R in (v/R) with r and plugging it in got 12.002 kg which is a very reasonable answer. I understand the need to account for small radius of the axle; however, I'm not sure I see the clear connection. Why would it not matter which R I replace with r?

10. Jul 17, 2014

### SammyS

Staff Emeritus
It's still true that I = (1/2) M R2. Right?

Really, unless you show the complete set of modified equations, it's difficult to comment on your updated result. Also, difficult to answer your most recent question definitively.

11. Jul 18, 2014

### onelove8187

Here is where I am at:

Still am attempting an energy approach so E_pot=E_k+E_rot

I still have the same acceleration I calculated from the falling mass being dropped to the floor where
t=10.42s
y_i=.64m

so: a_y=(2y_i)/t^2 gives a_y=-.0118m/s^2

and the the final velocity of the falling block (as well as V_t for rotating axle) is then
v_f=√(2*a_y*y_i) gives v_f=.1229

E_pot=mgy
E_k=.5mv^2

and E_rot=.5Iω^2
ω=(v/r_axle) and I=.5M(R_disk)^2

so E_rot=.5(.5MR_disk^2)(v/r_axle)^2

giving mgy=.5mv^2+.5(.5MR_disk^2)(v/r_axle)^2

solving for M gives (4mr^2(gy-(v^2)/2))/R^2V^2

the other required experimental data is
R_disk=.2282m
r_axle=.0192m
mass of falling block= 1.022kg

so M=(4*1.022*0.0192^2(9.8*.64-(0.1229^2/2)))/(0.2282*0.1229^2) gives M=12.00

This seems to give me a reasonable answer and it's the only way I could think to incorporate the differences in velocity. I have tried numerous times to approach this from a torque perspective but am struggling to wrap my mind around it. What is the best way to solve this with torque?

12. Jul 19, 2014

### SammyS

Staff Emeritus
It's helpful to have these details.

That method does give the correct final velocity.

Notice that putting these two equations together gives to following.
$\displaystyle v_f=\sqrt{2\left(\frac{2y_i}{t^2} \right)y_i\ }=2\frac{y_i}{t}\ .$​

This is consistent with $\displaystyle \ v_\text{avg}=\frac{v_i+v_f}{2}\,,\$ since $\displaystyle \ v_\text{avg}=\frac{y_i}{t}$

(Using this directly gives vf = 0.12284... m/s , which is close to your result .)
Doing it with torque should give the same result (of course). Use a free body diagram for the hanging mass.

In this approach you don't need velocities, but do need the acceleration. Your calculation of ay should be adequate here.