What is the angular acceleration of this cylindrical system?

[Np14]

Homework Statement

See picture.

Relevant Equations

τ[SUB]NET[/SUB] = Iα = F[SUB]NET[/SUB]r
a[SUB]T[/SUB] = rα
F = ma

PART B ONLY:
The cylinder undergoes torque when the mass m2 is removed:

τ_NET = Iα = F_NETr
= 45α = F_T(0.5)
F_T = 90α, therefore m_system = 90 kg

After this step, I am not sure what to do.

τ = ΣF = F_g - F_t
= ma = (20 kg)(9.8 m/s²) - (90α)
= 196 - 90α
α = 2.17 rad/s², which is incorrect

Can someone please help me figure out my mistake?

 

[Doc Al]

Realize that a and α are related but not the same.

 

[Np14]

Realize that a and α are related but not the same.

Yes I understand that. a_T = rα

 

[Doc Al]

Yes I understand that. a_T = rα

And you made use of that fact? (If so, I'll do the calculation myself and see what I get.)

 

[Np14]

And you made use of that fact? (If so, I'll do the calculation myself and see what I get.)

I did not use that equation, but I got the wrong answer, so it is most likely necessary to solve the problem. I'm pretty sure I did, however, differentiate between the two variables.

 

[Doc Al]

F_T = 90α, therefore m_system = 90 kg

What do you mean by m_system ?

 

[Np14]

I meant the mass of the two cylinders combined, which I thought constituted the entire system, but now I realize I didn't factor in the hanging weight, m1. Regardless I don't think that solving for the 90α (or the mass of the entire system) is relevant to the problem.

 

[Doc Al]

Start over with two equations: (1) For the torque on the cylinders & (2) For the net force on the hanging mass. (Use a = rα to relate the linear acceleration in (2) to the angular acceleration in (1).)

Solve these together. I suggest doing it symbolically, only plugging in numbers at the end.

 

[Np14]

Doing it symbolically makes it more confusing for me, so I will use numbers.

Torque on cylinders:
τ = Iα = ΣFr_perp.
τ = 45α = ΣF(0.5)
ΣF = 90α

Net force on hanging mass:
ΣF = F_g - F_t
= 20(9.8) - 20(a)
= 196 - 20(αr)
= 196 - 20(α(0.5))

Combine:
90α = 196 - 20(α(0.5))
100α = 196
α = 1.96 rad/s²

Thanks I got it!

 

[Doc Al]

Cool!

 

[Np14]

Doing it symbolically makes it more confusing for me, so I will use numbers.

Torque on cylinders:
τ = Iα = ΣFr_perp.
τ = 45α = ΣF(0.5)
ΣF = 90α

Net force on hanging mass:
ΣF = F_g - F_t
= 20(9.8) - 20(a)
= 196 - 20(αr)
= 196 - 20(α(0.5))

Combine:
90α = 196 - 20(α(0.5))
100α = 196
α = 1.96 rad/s²

Thanks I got it!

 

[Doc Al]

The way I would do it symbolically is like so (where I use T = tension):
(1) $Tr = I\alpha$
(2) $mg - T = ma = mr\alpha$

Multiply (2) by r:
(2') $mgr -Tr = mr^2\alpha$

Add (1) and (2'):
$mgr = I\alpha + mr^2\alpha$

Solve for $\alpha$:
$\alpha = \frac{mgr}{I + mr^2}$

Now plug in the numbers!