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The problem is as follows:

The figure from below shows two blocks joined by a rope on each end goind through a pulley. The masses of each block are as follows $m_1=5\,kg$, $m_2=4\,kg$. The coefficient of friction is $\mu=0.2$. The pulley can be considered as a thin disk and its mass is $2\,kg$. The acceleration due gravity is $10\,\frac{m}{s^2}$. Given this information find the acceleration in meter per second square of the system.

Sketch of the problem

The alternatives are as follows:

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

What I attempted to do to solve this problem was to relate the torque rotational inertia principle as follows:

$\tau=I\alpha$

This is illustrated in the diagram from below.

Sketch of the attempted solution

Since I'm given the masses of the objects and the acceleration due gravity and the coefficient of friction, then it is possible to obtain the acceleration as follows:

In this situation the tension acting in the pulley will generate a torque on it, hence.

For the block in the top:

$T-\mu N=ma$

$T=5a+0.2\times 50$

$T=5a+10$

For the block hanging from the other end in the pulley:

$T-mg=m(-a)$

$T=40-4a$

The rotational inertia for a disk with an axis going through the center is:

$I=\frac{1}{2}mr^2$

Therefore: (considering that a counterclockwise turn is positive and counterclockwise is negative)

$(R)(5a+10)-(R)(40-4a)=\frac{1}{2}mR^2 \times \alpha$

Simplifying:

$5a+10-40+4a=\frac{1}{2}(2)R \times \alpha$

Since the angular acceleration times the radius is the tangential acceleration this can be expressedas follows:

$9a-30=a$

$8a=30$

Therefore the acceleration should be:

$a=\frac{30}{8}$

Howevet this doesn't check with any of the alternatives, supposedly the answer to this question is $3\frac{m}{s^2}$ What could I be doing wrong? Can somebody help me with this matter?.

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  • $\begingroup$ The correct answer is $6 \frac{m}{s^2}$. $\endgroup$
    – YNK
    Dec 30 '19 at 15:18
  • $\begingroup$ $1.2\cdot 1.25= 1.5$ which is the ratio of the pulley and gravity accelerated blocks mass together, to the latter. $\endgroup$
    – user645636
    Dec 30 '19 at 16:57
  • $\begingroup$ You took the angular acceleration to be in counterclockwise direction when it will actually be in clockwise direction. The $I{\alpha}$ term becomes $-a$. $\endgroup$
    – Sam
    Dec 30 '19 at 18:17
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As Sam pointed out, the pulley will not roll without friction between the rope and the pulley. Hence the options are to consider 1) a system where they pulley doesn't rotate and the rope slides over it or 2) one where you assume there is enough friction between the pulley and rope such that no sliding occurs. The second case seems the most reasonable to me based on the problem statement. And this is what your solution method assumes. However, based on this assumption, the horizontal and vertical tension forces should be assumed to be unequal in your free body diagram. I.e. you should have a T1 for the horizontal force and a T2 for the vertical force. Nevertheless, your equation for the rotational acceleration is right, except that your sign for the acceleration on the right hand side of the equation should be negative (since the rotation will be clockwise, but you assumed counterclockwise was positive). Which gives $a=3m/s^2$.

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The pulley will not roll without friction on the surface of the pulley itself. From your free body diagram of the pulley, you can clearly see that net torque is zero. Hence applying ${\tau}=I{\alpha}$ does not make any sense. The acceleration of the string can easily be calculated and will be $a={30\over 9}m/s^2$

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  • $\begingroup$ The problem lies that such number is not the answer, hence my question. $\endgroup$ Dec 30 '19 at 14:24
  • $\begingroup$ The angular acceleration of the pulley will be zero because there is no torque on it. $\endgroup$
    – Sam
    Dec 30 '19 at 14:32
  • $\begingroup$ The angular acceleration of the pulley will be zero, assuming that the two tensions are in fact equal. (They shouldn't be.) $\endgroup$ Dec 30 '19 at 17:37
  • $\begingroup$ Well then the question should mention that the pulley has friction and rolls with no slipping to allow for that. $\endgroup$
    – Sam
    Dec 30 '19 at 17:43
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The balance of forces and torques and acceleration is actually given by the following sketch

Pulley_1

Therefore $$ \eqalign{ & \left\{ \matrix{ T_{\,1} = m_{\,1} \,\left( {a + \mu \,\,g} \right) \hfill \cr T_{\,2} = m_{\,2} \,\left( {g - \,a} \right) \hfill \cr \left( {m_{\,2} \,\left( {g - \,a} \right) - m_{\,1} \,\left( {a + \mu \,\,g} \right)} \right) = m_{\,p} \,a/2 \hfill \cr} \right. \cr & \left\{ \matrix{ T_{\,1} = m_{\,1} \,\left( {a + \mu \,\,g} \right) \hfill \cr T_{\,2} = m_{\,2} \,\left( {g - \,a} \right) \hfill \cr a = {{\left( {m_{\,2} \, - \mu m_{\,1} \,} \right)} \over {\left( {m_{\,1} + m_{\,2} + m_{\,p} \,/2} \right)}}g \hfill \cr} \right. \cr & \left\{ \matrix{ T_{\,1} = m_{\,1} \,\left( {a + \mu \,\,g} \right) = {{25} \over {10}}g \hfill \cr T_{\,2} = m_{\,2} \,\left( {g - \,a} \right) = {{28} \over {10}}g \hfill \cr a = {{\left( {4\, - 1\,} \right)} \over {\left( {5 + 4 + 1} \right)}}g = {3 \over {10}}g \hfill \cr} \right. \cr} $$

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  • $\begingroup$ To sum up the issue: The two tensions on the pulley need not be identical (and shouldn't be, if the pulley is to accelerate along with the blocks). $\endgroup$ Dec 30 '19 at 17:15
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    $\begingroup$ right: if the moment of inertia of the pulley is not negligible, it needs a torque (tangential force) to get accelerated, at the same tangential acceleration as the rope (if there is no slippage). $\endgroup$
    – G Cab
    Dec 30 '19 at 18:57
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enter image description here $T_h$ and $T_v$ are the tensions of the horizontal and vertical parts of the string respectively. These two tension are not equal. The difference between them is the force that rotates the pulley. Let us assume that the acceleration of the 2-mass system is $a\space\frac{m}{s^2}$. Therefore, the angular acceleration of the pulley is equal to $\frac{a}{R}\space\frac{rad}{s^2}$. The friction force acting on the mass on the table is $10\space\frac{kg\space m}{s^2}$. The three equations needed to find $a$ is given below.

for the mass on the table:$\space\space T_h-10=5a$

for the hanging mass:$\space\space\space\space\space\space\space 40-T_v=4a$

for the rotating pulley:$\space\space\space\space\space \left(T_v-T_h\right)R=\frac{1}{2}2R^2\frac{a}{R}=\space\space\to\space\space T_v-T_h=a$

Add the three equations together to get $$30=10a$$ $$\therefore\space\space a=\frac{30}{10}=3\space\frac{m}{s^2}$$

$\underline {Note}$: I am sorry for posting earlier a wrong answer without giving any proof. I am posting this answer hoping to makes up for that. The answer given here tallies with the answer posted by $\bf {gigo318}$.

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