# Finding the mass of a rotating wheel with an attached mass

1. Jul 20, 2014

### Jesse2789

1. The problem statement, all variables and given/known data
Find the mass, M, of a rotating wheel of radius r that has an attached mass, m, suspended by a string using conservation of energy. The mass is suspended a height, h, above the ground and it takes a time of t seconds to reach the ground.

2. Relevant equations
$$U_g,mass = K_f,mass + K_f,wheel$$
$$mgh = 1/2(m_b)(v_f)^2 + 1/2(Iω_f^2)$$ (m_b is the hanging block's mass)
$$I_(disk) = 1/2(Mr^2)$$

3. The attempt at a solution

$$a_y,block = (-2h)/(t^2)$$
$$v_f,y,block = (a_y,block)*t$$

^At this point the only two unknowns in the conservation equation are I and ω. To find ω_f could I just say that v_f,y,block is the same as the final tangential velocity of the wheel?

So then it would just be $$ω_f=v_t/r$$, then you could find M through $$I=Mr^2$$?

EDIT: I just tried this with numbers and came out to an unreasonably high mass of 1879kg. I did this in lab and could lift the wheel, so obviously there's an error somewhere in my reasoning.

Last edited: Jul 20, 2014
2. Jul 21, 2014

### SammyS

Staff Emeritus
Hello Jesse2789. Welcome to PF !

Is the string wrapped around the circumference of the wheel, or is it wrapped around an axel of smaller radius?

What numbers did you use & where did you get them?

3. Jul 21, 2014

### ehild

Hi Jesse, welcome to PF!

You forgot to explain how that string is attached to the wheel. Is it wound on the rim?

Yes, it is right, but I = 1/2 Mr2.

What is your final equation for M? What were the data?

ehild

4. Jul 21, 2014

### Jesse2789

Values & calculation

Alright, so the string is spun around the axle first of all.

Data collected from Trial #1 is as follows: Time to fall = 10.33 seconds, distance fallen = 0.557 meters, radius of wheel/axle system is 0.237 meters.

I ended up with $$M=4(mgh-0.5mv_f^2)/(r^2ω^2)$$, and with values this came out to an incredible 1879kg.

Here's also some of the work that led up to my final answer:

$$a_y = -2h/t^2 = -2(0.557m)/(10.33sec)^2 = -0.01044 m/s^2$$
$$v_f,y,block = a_yt = (-0.01044 m/s^2)(10.33sec) = -0.1078 m/s$$
$$ω = v_t/r = (-0.1078 m/s)/(0.237m) = -0.4550 rads/sec$$
$$mgh = 1/2mv_f^2+1/2(1/2Mr^2)ω^2$$

And then plugging in all values I found M to be 1879. (I have a feeling you cannot say that the tangential velocity of the wheel is the same as the final velocity of the block)

--Sidenote: I voiced my difficulties I was having with my professor via e-mail, as I have the class tomorrow, and she gave me an extension on the due date. So, I will be checking back on this post in a few hours when I wake up. Cheers! :)

Last edited: Jul 21, 2014
5. Jul 21, 2014

### ehild

You did not give the value of m and the radius of the axle.

ehild

6. Jul 21, 2014

### Jesse2789

The mass was varied over a few runs but for the initial calculation it started with m=1.002kg and the radius of the wheel & axle system is 0.237m. I found out what I did incorrectly though..

I said that $$ω=v_t/r= -0.4550 rad/s$$ but then when I plugged ω back into that equation I said $$ω^2 = (v_t)^2/(r^2) = ((-0.4550 rad/s)/(0.237m))^2$$
^^Right here I should not have divided by the r^2.

The correct answer was based off of the work done by gravity and the work done by the tension of the string. I'll post the work below (my professor showed this to me after class today).

Tension of string $$T = -ma_y + mg = m(g-a_y)$$

Then using that tension applied to net work..

Net work = work done by tension - work done by gravity
$$W_(net)=W_T-W_w$$
This gives $$5.83 * 10^-3 J = W_(net)$$

$$5.83 * 10^-3 J = 1/2Iω^2$$
$$5.83 * 10^-3 J = 1/4Mr^2ω^2$$

Then using the the previously solved for $$ω=v_t/r$$ I solved the mass, M, to be about 2.01 kg.

Sorry that I wasn't clear enough originally, I'll make sure to be much more concise next time! Also thanks for the welcome :-), I've been a guest browsing through the forums for quite a while.

Last edited: Jul 21, 2014
7. Jul 22, 2014

### dean barry

You have the distance s and time t, so you can find the acceleration rate and final velocity

Find the linear KE of the block from ½ * m * v ²

The rotating KE = (m*g*s) - ( ½ * m * v ² )

The rotation rate of the wheel (ω) = v / r