# How should it be approached the combined moment of inertia of a system and torque to find the angular acceleration of a system of pulleys? [closed]

I'm not very sure if the combined system of pulleys sharing a common radius should it be approached if it would be a single one. In either case, what would be the way to assess that situation?.

To illustrate this I found this problem of an unknown author which has left me confused in the subject.

The problem is as follows:

The figure from belows 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 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 it is to find the values of each tension acting on the wheels.

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

$$T-mg=m(-a)$$

$$T=300-30a$$

For the block which has a less heavier 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?. Does this means that the combined moment of inertia should it be split between the two pulleys or what?

• Hi. Voting to close this question as valid homework-like questions do not include "check my work" type of questions. – Dvij D.C. Dec 30 '19 at 14:07
• @DvijMankad I did reworded the heading of this question. I hope it does meet the criteria as required by this stack community. – Chris Steinbeck Bell Dec 30 '19 at 14:22
• Hi Chris, the title wasn't the issue. As you say, you have only reworded the title--the content of the question has remained the same. Instead of showing the full calculation you have done, you should simply ask the conceptual question that you are struggling with--something like what you have mentioned in the last sentence of your question. But of course, you should elaborate a bit more and be more precise about what you mean by that. – Dvij D.C. Dec 30 '19 at 14:29
• Using your equations which imply positive clockwise rotation, I get the torque equation in this form: [300 – 30(.4)α](.4) – [200 + 20(1.2)α](1.2) = 26.4 α Note that one torque is CW and the other is CCW. This solves to α = -2 r/s/s. The left side torque is larger and the acceleration is CCW. – R.W. Bird Jan 1 '20 at 16:00

You try to calculate net force, instead you should calculate a net torque: $$m_1gR_1-m_2gR_2=I\alpha$$
Plunging in all the data and solving for $$\alpha$$, gives $$4.5 \space [ \frac{rad}{s^2} ]$$.