Solving the Puzzle: Forces & Acceleration

[jamesdubya]

I had just asked a question about boats going across a river but that question and this question really have me stumped.

Homework Statement

http://img138.imageshack.us/img138/7910/68512243.png
A)Find the acceleration using Newtons laws
B)Find the velocity using Conservation of Total Mechanical Energy.

Homework Equations

Idisk=1/2mR^2 (what is R?)
M1=T+Ff-MG=ma
M2=1/2mR^2-t+t=ma
M3=-T-Mg=ma
I doubt these are right but I tried, I'm terrible at summing forces.

The Attempt at a Solution

A)FnMuk-1.1(9.8)+1/2mR^2-t+t-t-Mg=m(a)
I know I am way off but just so you can see I at least tried.

 

[rl.bhat]

Radius of the disc R is required to find the acceleration.
The tensions in the two sections of the string are not equal.
So rewrite the equations for M1, M2 and M3.

 

[jamesdubya]

m1=T-MG-Fn-FF ?
m2= ?? My only guess would be the previous equation.
m3=T+MGH

 

[tiny-tim]

m1=T-MG-Fn-FF ?
m2= ?? My only guess would be the previous equation.
m3=T+MGH

No, you're ignoring what rl.bhat said …

there are two tensions, so call them T₁ and T₂.

(yes, I know the tension along a string is usually constant, but that's only if the pulley is frictionless … in this case, there is friction at the pulley, otherwise it wouldn't turn, so the tension is different on either side.)

 

[jamesdubya]

ohh ok i see i see, thank you for your response... so
M1=Muk*M+T1
M2=M*.5mr^2+T1-T2
M3=T2-mg
any better?

 

[tiny-tim]

ohh ok i see i see, thank you for your response... so
M1=Muk*M+T1
M2=M*.5mr^2+T1-T2
M3=T2-mg
any better?

(have a mu: µ and try using the X₂ tag just above the Reply box )

You're not writing these properly.

The first and third equations are F = ma.

The second equation is torque = Iα.

Start again (and be careful about the ±s). ​

 

[jamesdubya]

M₁=$$\mu$$*M-T₁=ma
M₂=I$$\beta$$
M₃=-T₂+MG=ma
any better?

 

[tiny-tim]

M₁=$$\mu$$*M-T₁=ma
M₂=I$$\beta$$
M₃=-T₂+MG=ma
any better?

(what happened to that µ i gave you? )

The third equation is correct.

In the first equation, the signs are wrong.

The second equation isn't an equation (and what is ß anyway?)

 

[jamesdubya]

I tried to use it! it came out goofy looking.. By the big B i ment to put$$\alpha$$.
M₁=-µ*M+T₁=ma
M₂= I $$\alpha$$= M₂G+T₂
M₃=-T₂+MG=ma
what about now?

 

[tiny-tim]

By the big B i ment to put$$\alpha$$.

M₁=-µ*M+T₁=ma
M₂= I $$\alpha$$= M₂G+T₂
M₃=-T₂+MG=ma
what about now?

The first equation is fine now (and easy to read ), except you need a g (why are you writing "G"?)

In the second equation, why do you have an Mg … that pulley isn't going anywhere, is it? Also, you need to change α into a function of a, you need to use T₁, and you need torque on the RHS, not force.

 

[jamesdubya]

ok so then
M₂=M₂+T₁=I$$\alpha$$
I don't really understand what you mean change a to a function of a sorry. Does this look at least a little better

 

[tiny-tim]

ok so then
M₂=M₂+T₁=I$$\alpha$$
I don't really understand what you mean change a to a function of a sorry. Does this look at least a little better

grrr … not much.

What is M₂ doing there?

What happened to T₂?

And, I repeat, T is a force, you need a torque

(and you need a in all three equations, or you won't be able to solve them , so you must write α in terms of a)

 

[jamesdubya]

m2 is just a pulley it is just letting the rope go through it pretty much. So add a -T₂ and set it = r x F which is =r x ma ?

 

[tiny-tim]

m2 is just a pulley it is just letting the rope go through it pretty much. So add a -T₂ and set it = r x F which is =r x ma ?

I think you need to get some sleep. :zzz:

Start again in the morning. ​

 

[jamesdubya]

I fell asleep at 6 and woke up super early, I am up for the day.Can you please just help me with this I just am not understanding how to do torque..

 

[tiny-tim]

I fell asleep at 6 and woke up super early, I am up for the day.Can you please just help me with this I just am not understanding how to do torque..

Does that mean you've only had about 2 hours sleep?

You've been making silly mistakes for the last three posts. If you're not normally like this, then you need more sleep.

torque is force times (perpendicular) distance, in this case tension times radius.

 

[jamesdubya]

no i had about 8 slept from 6-230ish...anyway...I just don't understand it, they may be silly to you but this makes 0 sense to me, that's why I came here, I wouldn't be here if i knew how to do it well.. I do appreciate your attempts at help though.. ok so then M2=t₂*radius ?

 

[tiny-tim]

M2=t₂*radius ?

No, torque = Iα, so Iα = T₂*radius - T₁*radius.

ok, now what is α in terms of a?

 

[jamesdubya]

a\R=$$\alpha$$?

 

[tiny-tim]

a\R=$$\alpha$$?

Yup!

ok, now put all the numbers into the three equations, and solve them. ​

 

[jamesdubya]

M1=-µ*Mg+T1=ma
m2 = T2*radius - T1*radius=Iα
M3=-T2+Mg=ma

µ*M₁g+T₁-T₂+M₃g=ma
.36*1.1*9.81+5*9.81=1.6(a)
33.08=a
is that right?
I don't know how to find the radius for the disk

 

[tiny-tim]

µ*M₁g+T₁-T₂+M₃g=ma
.36*1.1*9.81+5*9.81=1.6(a)
33.08=a
is that right?

Is M₃ 5 kg or 0.5 kg?

And you left out a minus.

I don't know how to find the radius for the disk

You don't need it, it cancels.

You need to eliminate T₁ and T₂ (because you don't know them, and the question doesn't ask for them), so get two equations for T₁ - T₂.​

 

[jamesdubya]

It's 5kg.
t₁-t₂=Iα
and then I had a t₁-t₂ in my other equation that would cancel these out.