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.
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.
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?.
Best Answer
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 $a=3m/s^2$.