# Find mass of cylinder (rotational motion question)

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1. Mar 1, 2015

### Valerie Prowse

1. The problem statement, all variables and given/known data
1. A light string is wrapped around a solid cylinder and a 300 g mass hangs from the free end of the string, as shown. When released, the mass falls a distance 54 cm in 3.0 s.

a. Draw free-body diagrams for the block and the cylinder. b. Calculate the tension in the string.

c. Calculate the mass of the cylinder
2. Relevant equations
τ = rF
I = 1/2mr^2
mg - Ft = ma
τ = Iα

3. The attempt at a solution

I'm pretty sure I have figured out part a) and b) (tension in the rope = 2.9 N), but I'm not sure about part c).

I know that r needs to be cancelled out at some point, because we are not given the radius of the cylinder. I am thinking it goes something like this, but I get lost somewhere in here:

F = Force of tension in the rope {and, from part a., F = m(g-a) }
M = mass of cylinder
m = 3.0 kg
α = a = 0.12 m/s^2

τ = rF = Iα

rF = 1/2Mr^2 * α
r (m[g-a]) = 1/2Mr^2 * α (this is where I get lost, I'm not sure what to do)
2rm(g-a) / r^2 * α = M

I'm not sure if this is the right thing to do... I'm not sure how I could make sure that r cancels out. Or maybe it is the wrong equation to use in general. Any help in the right direction is appreciated! :)

2. Mar 1, 2015

### Suraj M

You can get the velocity $v$ here, which is same as the linear velocity of the particle on the surface of the cylinder.
so here get $\omega$ in terms of radius.
also from the above data you can calculate the change in potential, the kinetic energy gained by the 300 g object, they should have been equal but a part the potential energy gets converted into rotational energy of the cylinder. So find the difference between PE and KE and equate to your formula for rotational energy!
as your $\omega$ is in terms of r-2 and as I(moment of inertia) is in terms of r² you can eliminate your radius.

3. Mar 1, 2015

### ehild

You need to know how the linear acceleration a and the angular acceleration α are related.
When the hanging box travels y distance, the length of the hanging part of the rope increases by y. That length s=y has to be unwrapped from the cylinder, so as it turns by the angle θ=s/r.
Accordingly, the velocity of the rim is v=rω and you get the the analogous relation between the acceleration a and angular acceleration α.

4. Mar 1, 2015

### Valerie Prowse

So, I found velocity to be 0.36, which means that the linear velocity is also 0.36, but how do you get ω without r?
I did the same thing with the kinematics for rotational motion:
ω = ωo + αt = 0.36 rad/s, but I don't feel like this is right, and that α does not = a.

5. Mar 1, 2015

### Suraj M

Don't worry, introduce r, it'll get cancelled out later. $v=r\omega$ so get $\omega$ in terms of v and r.
Why do this when you can do this..
the v you found out, use it find the KE of the body,
also, calculate the Rotational energy of the cylinder: $RE=½I \omega^2$ ....$I=½mr^2$ and $\omega^2 = \frac{(0.36)^2}{r²}$ your r will get cancelled leaving you with mass, which you require, and as i said earlier, equate this rotational energy to difference in change in potential and KE (at 3 seconds). It will give you an answer!

6. Mar 1, 2015

### Valerie Prowse

Ah okay, I see what you mean. So, the distance travelled by the block has to = the distance travelled in radians around the cylinder (which is θ = L/r, the length between θo and θ divided by the radius)? Does this mean I should figure out a way to incorporate equations about the block's movement and equations about the cylinder's movement, since they are directly related? (Sorry, in addition to being poor at math, I'm just trying to understand conceptually what is happening as well!)

If this is the case, is it very important which equations about the motion I should be using, or does it not matter since the motion of the block and cylinder are related in multiple ways?

7. Mar 1, 2015

### Valerie Prowse

The reason I am avoiding this is because we haven't yet been introduced to rotational energy (next lesson), so I am unfamiliar with how to use the information. But I appreciate the help! :)

8. Mar 1, 2015

### Suraj M

I understand :)... so you can try this:
the tension in the rope causes a torque, $t=r×F$ where F is the tension,
this torque causes a change in angular momentum$$t =r×F= \frac{I\omega_{final}- I \omega_{initial}}{time}~~~ \left(variant~ of ~newtons ~second ~law ~of ~motion\right)$$
$I_{initial}=0$
as you know $\omega$ in terms of r, substitute , r will get cancelled, giving you mass

Last edited: Mar 1, 2015
9. Mar 1, 2015

### PeroK

I like this equation you've found. It was exactly the right thing to do. Although you should have brackets round $(r^2 \alpha)$.

All you need to do is calculate $a$ calculate $\alpha$ (which is related to a), plug them in and you have a nice formulaic answer. You don't need to start calculating angular velocity or rotational energy. You were very close with that equation.

10. Mar 1, 2015

### Valerie Prowse

How is this? If α = a/r, I can use that to plug in to α in the equation?

if so, I THINK I have it:

2rm(g-a) / (r^2 * α) = M
2rm(g-a) / (r^2 * [a/r]) = M
which becomes
2r^2m(g-a) / (r^2 * a) = M
and cancel out the r^2
2m(g-a) / a = M

M = 484kg
Although that number seems way too large for the context of the question..

EDIT: My bad, I put in m = 3.0kg instead of 0.3.
M = 48.4 kg

11. Mar 1, 2015

### PeroK

12. Mar 1, 2015

### Valerie Prowse

Thank you everyone for your help!!

13. Mar 1, 2015

### ehild

Correct, well done!

14. Apr 11, 2015

### SteveS

I just completed this question as well. I'm glad to see i got the same answer but with a different approach for part c.

i used the formula

t = la
l= MR^2/2
a = a/r

t = MRa/2

then i Knew that my F on the cylinder from the free body diagram was 2.904 N in the downward direction.

t = r x F

Equating these two equations to each other cancels out the R and leaves mass as the unknown.

M = 2 ( -2.904) / -0.12 = 48.4 kg

Seems a bit cleaner than the above approach.

15. Dec 10, 2016

### bruuuuuuhhhhhhhhhhh

Would the Velocity not be .54/ 3 = .18m/s. Then To calculate the acceleration divide by 3 = .06 m/s^2. From there it is correct however the mass would be double?

Or am I a fool?

16. Dec 10, 2016

### TomHart

Hi bruh, welcome to Physics Forums.
You can find the average velocity by dividing 0.54/3 = 0.18 m/s. But to then divide that by 3 to get the average acceleration is not valid.
For constant acceleration, velocity with respect to time is linear. So with an initial velocity of 0 m/s, for the average velocity to be 0.18 m/s over a 3 second time period, the final velocity at 3 seconds would have to be 2 times 0.18 m/s, or 0.36 m/s. And the velocity at the midpoint time (t = 1.5s) would be 0.18 m/s.

So the average acceleration would be 0.36/3 = 0.12 ms-2

17. Apr 25, 2017

### chemtal

im still confused with part B. my attempt:
sum of F in the mass= ma=mg-Ft
mg-ma=Ft
(0.3kg)(9.8m/s^2)-(0.300kg)(a)=Ft
But what is a? im assuming its not "g" because the tension would decrease the acceleration. Also, do we need to take friction into account?

18. Apr 25, 2017

### chemtal

O i figured it out! i used y=y0+v0t=(1/2)at^2 and got an acceleration of 0.12m/s

19. Apr 25, 2017

### chemtal

so i got the 48.4kg=M, but is this not the total mass of the system, so wouldnt the mass of the cylinder be M-m=48.4-0.3-48.1, which is still 48 when you round it to two significant digits, but i just want to make sure that I am understanding where all the variables are coming from. or am i mistaken an M only represents the mass of the cylinder?

20. Apr 25, 2017

### TSny

Why do you think M is the total mass of the system? What equation(s) did you use to get the answer for M?