How Is Angular Acceleration Derived in a Pulley System?

[Linus Pauling]

1.

Find magnitude of angular accelleration alpha, express in terms of radius r and gravity



2. torque = rF, F=ma



3. T = mg
torque = rF = -Tr = I*alpha
-mgr = I*alpha
alpha = -mgr/I

with I = 0.5 mr^2 I get:

alpha = -2g/r, which is incorrect.

 

[Tzim]

Why t=ma.It is not stationary.it is moving.so mg-T=ma

 

[Tzim]

The body is moving with an acceleration of a which is alpha*r

 

[Linus Pauling]

I'm still not getting the solution.

Using ma = mg - T
I*alpha = -Tr
a = r*alpha

I obtained:

alpha = (2a - 2g)/r

Using ma - mg = I*alpha/r I obtained:

a = 0.5*r*alpha + g

Putting a into the equation for alpha:

2[(0.5*r*alpha + g) - g]/r
a = alpha

 

[Tzim]

I don't really get what you did.You don't have to put - at the equation Tr=I*alpha.Also alpha=a/r.try this

 

[Linus Pauling]

I*alpha = -Tr because T exerts torque in the negative (clockwise) direction.

 

[Tzim]

Nevertheless if you put numbers you would found the same.The idea though is the one i wrore you.i hope i helped.if you have more questions just ask

 

[Linus Pauling]

Sorry, this is not helping. Using, alpha = Tr/I and I = 0.5mr^2, I obtained:

alpha = [m(g-a)]/(0.5mr^2)
Solving after quite a bit of algebra:

alpha = g/(r^2 + r) is incorrect

 

[Tzim]

Which is the correct answer?Try to solve as to T.T=-I*alpha/r
and also write alpha as a/r.then put T at the equation mg-T=ma

 

[Tzim]

is the correct answer 2g/3r?

 

[Linus Pauling]

...... I don't know the correct answer obviously.

I have done all that already. With T = -I*alpha/r, alpha = a/r, and T = mg - ma, I obtain:

alpha = [rm(g-a)]/-I = (2a - 2g)/r

 

[Linus Pauling]

is the correct answer 2g/3r?

Yes, how did you calculate this?

 

[Tzim]

How you obtain all this?-Tr=0.5mr^2*a/r so T=-0.5ma
Then mg -T=ma so mg-0.5ma=ma so 3/2*a=g so a=2g/3