Angular momentum and acceleration

[Felafel]

Homework Statement

m1 and m2 are two blocks tied with a rope with a pulley inbetween, like those in this picture
http://labella.altervista.org/images/mechanicsoftwopoint_2.png
there are no frictions.
find: the linear acceleration of the blocks and the tension of the rope on both m1 and m2.

The Attempt at a Solution

I'd say it is:
$m_2 \cdot g = I \alpha + m_1 \cdot a + m_2 \cdot a$ with $a= \alpha \cdot R_0$ being $R_0$ the radius of the pulley
$m_2 \cdot g= m_1 \cdot a + I \frac{a}{R_0}+m_2 \cdot a$ so
$a=\frac{m_2 \cdot g}{mb + \frac{I}{R_0}}$

and then the tension of the rope on m1 is simply
T=a*m1
while the tension on m2 is
T=m2*g+m2*a

Is it correct? If not, why?
Thank you for checking :)

 

[Doc Al]

Does the pulley have mass?

 

[Felafel]

it isn't written, but I don't think so.

 

[Doc Al]

it isn't written, but I don't think so.

Then you can forget about I for the pulley.

Apply Newton's 2nd law to each mass.

 

[Felafel]

okay, thank you. :)
But if the pulley had mass, would it be right?

 

[Doc Al]

okay, thank you. :)
But if the pulley had mass, would it be right?

No, it's not correct.

If you want to know why it's wrong, show how you got that result.

 

[Felafel]

i thought $m_2 \cdot g$ is the force which moves the system, so it equals
$m_1 \cdot a$ with an unknown acceleration, $+m_2 \cdot a$ because the block goes down with the same acceleration, and it also makes the pulley spin with $I \cdot \alpha$
then I've read the formula $a= \alpha \cdot R_0$ on a book, and thought of applying it

 

[Doc Al]

i thought $m_2 \cdot g$ is the force which moves the system, so it equals
$m_1 \cdot a$ with an unknown acceleration, $+m_2 \cdot a$ because the block goes down with the same acceleration, and it also makes the pulley spin with $I \cdot \alpha$
then I've read the formula $a= \alpha \cdot R_0$ on a book, and thought of applying it

Rather than try to do it all at once, and risk making an error, why not divide and conquer. Apply Newton's 2nd law to the two blocks and to the pulley.

 

[Felafel]

like:
$F_g= m_2 \cdot g$
$F_1= m_1 \cdot a$
$F_p= I \cdot \alpha$
$F_2= F_g-F_1-F_p$
?
i keep getting the same result

 

[Doc Al]

$F_1= m_1 \cdot a$

This one I understand. Where $F_1$ is the tension in string 1.

$F_p= I \cdot \alpha$
$F_2= F_g-F_1-F_p$

These I don't understand.

What forces act on $m_2$?
What forces act on the pulley? What torque do they exert?

 

[Felafel]

on m2 we have the gravitational force:
$F_p=m_2 \cdot g -T= m_2 \cdot a$ where T is the tension of the string
$T= m_1 \cdot a$
by subtracting the two:
$m_2 \cdot g - m_1 \cdot a = m_2 \cdot a$
and $I \cdot \alpha= m_2 \cdot g - T$
$I \cdot \alpha= m_2 \cdot g - m_1 \cdot a$
but i really don't know what I am doing :(

 

[sankalpmittal]

on m2 we have the gravitational force:
$F_p=m_2 \cdot g -T= m_2 \cdot a$ where T is the tension of the string
$T= m_1 \cdot a$
by subtracting the two:
$m_2 \cdot g - m_1 \cdot a = m_2 \cdot a$
and $I \cdot \alpha= m_2 \cdot g - T$
$I \cdot \alpha= m_2 \cdot g - m_1 \cdot a$
but i really don't know what I am doing :(

If the pulley had mass, the tension at its two ends shouldn't be same !

In fact,

(T₁-T₂)*r= Ia/r

T₁ is that downward tension, and T₂ leftwards.
You need to get two more equations by applying Newtons second law on each mass for translational motion.

Approach 2 : Use conservation of mechanical energy on the two blocks system.

 

[Felafel]

Thank you :)!
but I don't understand how you found the left part of this equation: (T1-T2)*r= Ia/r
i mean, why is the difference between the two tensions multiplied by r?

 

[Saitama]

but I don't understand how you found the left part of this equation: (T1-T2)*r= Ia/r
i mean, why is the difference between the two tensions multiplied by r?

Torque is a vector. Determine the directions of torques produced by tensions in the rope.

 

[Doc Al]

but I don't understand how you found the left part of this equation: (T1-T2)*r= Ia/r
i mean, why is the difference between the two tensions multiplied by r?

When applied to rotation, Newton's 2nd law becomes:
Ʃ Torque = Iα

Again, start by answering the questions I had asked earlier. What forces act on the pulley? What torques do they exert? Then you can apply Newton's 2nd law to the rotation of the pulley.

 

[Felafel]

i'd say $T_2=m_2 \cdot g$ and $T_1=M_1 \cdot a$ are the only forces acting on the pulley
and so $\Sigma \tau = (T_1- T_2)x \vec{r}$ where r is the motion (which equals the radius?). which gives
$(m_2 \cdot g - m_2 \cdot a) x r = I \cdot \frac{a}{r}=$
$a=\frac{m_2 \cdot g}{\frac{I}{r^2} + m_1}$ right?

 

[Doc Al]

i'd say $T_2=m_2 \cdot g$ and $T_1=M_1 \cdot a$ are the only forces acting on the pulley

Yes, the two tensions are the only relevant forces acting on the pulley. But your first equation, for block 2, is incorrect. (If $T_2=m_2 \cdot g$ then the net force on block 2 would be zero.)

and so $\Sigma \tau = (T_1- T_2)x \vec{r}$ where r is the motion (which equals the radius?). which gives
$(m_2 \cdot g - m_2 \cdot a) x r = I \cdot \frac{a}{r}=$
$a=\frac{m_2 \cdot g}{\frac{I}{r^2} + m_1}$ right?

Do it systematically. For the pulley:

$\Sigma \tau = (T_1 - T_2) \cdot R = I \cdot \frac{a}{R}$

That's one equation. Now write equations for each of the blocks. (The equation you wrote for block 1 is correct; fix the one for block 2.)

 

[Felafel]

is block 2:
$F= -m_2 \cdot g + m_2 \cdot a$?

 

[Doc Al]

is block 2:
$F= -m_2 \cdot g + m_2 \cdot a$?

Please use the same notation you used earlier. What's F supposed to be?

What forces act on block 2? What are their directions?

 

[Felafel]

oops, sorry
then, on block two there should be the gravity force:
$-m_2 \cdot g$ and the tension of the string: $T_1$ which have opposite directions. then the block moves downwards with an acceleration = "a"
so:
$T_2-m_2 \cdot g =- m_2 \cdot a$

 

[Doc Al]

oops, sorry
then, on block two there should be the gravity force:
$-m_2 \cdot g$ and the tension of the string: $T_1$ which have opposite directions. then the block moves downwards with an acceleration = "a"
so:
$T_2-m_2 \cdot g =- m_2 \cdot a$

Good. Now you have all three equations. Solve for the acceleration.