How to find the angular acceleration of a system of pulleys interconnected?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

The figure from below shows a system featuring two coupled pulleys have $R_1=1.2\,m$ and $R_2=0.4\,m$, The mass of $m_1=20\,kg$ and $m_2=30\,kg$ and the total moment of inertia of the two pulleys is $26.4\,kg \cdot m^2$ . Find the angular acceleration in $\frac{rad}{s^2}$ of the system of pulleys. The acceleration due to gravity is $g=10\,\frac{m}{s^2}$.



The alternatives are as follows:

$\begin{array}{ll}
1.&10\,\frac{rad}{s^2}\\
2.&8\,\frac{rad}{s^2}\\
3.&6\,\frac{rad}{s^2}\\
4.&4\,\frac{rad}{s^2}\\
5.&2\,\frac{rad}{s^2}\\
\end{array}$

What I've attempted to do is displayed in the sketch from the figure from below:



After identifying the forces acting on the pulleys, I could spot that the best approach would be to find the torque acting on the wheels and from then finding the angular acceleration as follows:

$\tau=I\alpha$

In a similar fashion to that of the Atwood's machine, what I did was the following, which is to find the values of each tension acting on the wheels.

For the block which mass is the heavier of the two:

$T-mg=m(-a)$

$T=300-30a$

For the block which has a less heavy mass:

$T-mg=ma$

$T=mg+ma$

$T=200+20a$

Therefore:

$-(1.2)\left(200+\frac{20\alpha}{1.2}\right)-(0.4)\left(300-\frac{30\alpha}{0.4}\right)=26.4\alpha$

Solving this equation results into:

$\alpha\approx -21.95$

But this answer doesn't get any close to what the alternatives do say. Supposedly, this answer results into $2\frac{rad}{s^2}$, but I'm confused because each time I go back in my steps I end up getting the same answer. Can somebody help me with this please?
 

skeeter

Math Team
Jul 2011
3,356
1,848
Texas
$m_1g-T_1 = m_1a_1 \implies T_1 = m_1g-m_1r_1\alpha$

$T_2 - m_2g = m_2a_2 \implies T_2 = m_2g + m_2r_2\alpha$

substitute the expressions for $T_1$ and $T_2$ into the net torque equation and solve for $\alpha$ ...

$T_1r_1 - T_2r_2 = I\alpha$
 
Jun 2017
399
6
Lima, Peru
$m_1g-T_1 = m_1a_1 \implies T_1 = m_1g-m_1r_1\alpha$

$T_2 - m_2g = m_2a_2 \implies T_2 = m_2g + m_2r_2\alpha$

substitute the expressions for $T_1$ and $T_2$ into the net torque equation and solve for $\alpha$ ...

$T_1r_1 - T_2r_2 = I\alpha$
I'm revisiting this problem. Can you please explain why?

$m_1g-T_1 = m_1a_1$

and

$T_2 - m_2g = m_2a_2$

The way how I see it is the opposite.

The mass of the left side has lower mass. So it should go upwards, while the one which is on the right ride has bigger mass and will go downwards.

Given this situation it would be:

$T_1 -m_1g = m_1a_1$

and

$m_2g -T_2 = m_2a_2$

as in this situation the acceleration is negative.

Can you include a FBD for this thing?. I'm still confused about this problem. Help!
 

skeeter

Math Team
Jul 2011
3,356
1,848
Texas
The way how I see it is the opposite.

The mass of the left side has lower mass. So it should go upwards, while the one which is on the right ride has bigger mass and will go downwards.
Answer this question ... Leaving the 30 kg mass at $m_2$, how much mass is required at $m_1$ to make the system balance (i.e. equilibrium).
 
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Jun 2017
399
6
Lima, Peru
Answer this question ... Leaving the 30 kg mass at $m_2$, how much mass is required at $m_1$ to make the system balance (i.e. equilibrium).
That would be $10\,kg$. Isn't it?. Am I right with this calculation?. The torque provided by one end (the right side) is $120 Nm$ thus to compensate this would need that torque in the other side of the pulley. Is this the reason why that side (contrary to a misguide due looking solely to the masses) will go downwards?.