How to find the acceleration of a pulley when two masses are held to it?

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

A pulley has a mass of $1.2\,kg$ and radius equal to $5\,cm$ has two bobs hanging from it as indicated in the figure from below $m_1=0.8\,kg$ and $m_2=0.6\,kg$ from both ends. The wire has a negligible mass and passes through the cavity of the pulley. The friction between the wire and the pulley allows that it moves when both masses moves. Given these conditions find the acceleration in meters per second square for the masses. Assume $g=10\,\frac{m}{s^2}$



The alternatives are as follows:

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

I'm confused exactly on what way should I assign the signs for this object. I'm currently getting $10\,\frac{m}{s^2}$. But I don't know if is it because I'm assuming that the tension for each side is different or what?. Can someone help me with this please?.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
$a=10 \, m/s^2 \implies$ a free fall condition, which does not make sense.



Clearly, the left mass will accelerate downward and the right mass will accelerate upward ... have you ever been taught the use of scalar equations?

Let $a$ be the magnitude of the acceleration for both masses. Both equations are set up such that greater force minus lesser force yields a positive value.

$0.8g - T_1 = 0.8a$

$T_2 - 0.6g = 0.6a$

$(T_1 - T_2) \cdot R = I \alpha \implies T_1 - T_2 = \dfrac{1}{2}M_p \cdot a$

summing the three equations term for term yields ...

$0.2g = 1.4a + 0.6a \implies a = 0.1g = 1 \, m/s^2$
 
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Jun 2017
399
6
Lima, Peru
$a=10 \, m/s^2 \implies$ a free fall condition, which does not make sense.



Clearly, the left mass will accelerate downward and the right mass will accelerate upward ... have you ever been taught the use of scalar equations?

Let $a$ be the magnitude of the acceleration for both masses. Both equations are set up such that greater force minus lesser force yields a positive value.

$0.8g - T_1 = 0.8a$

$T_2 - 0.6g = 0.6a$

$(T_1 - T_2) \cdot R = I \alpha \implies T_1 - T_2 = \dfrac{1}{2}M_p \cdot a$

summing the three equations term for term yields ...

$0.2g = 1.4a + 0.6a \implies a = 0.1g = 1 \, m/s^2$
It seems that I made a mistake with the multiplication on the left side of the equation when comparing my notes. But I followed your system of equations. But the original source of my confusion is not yet answered. Why is it a different tension on the other side of the wire?. Isn't the tension the same when a wire goes through a pulley? Or am I missunderstanding the message?.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
But the original source of my confusion is not yet answered. Why is it a different tension on the other side of the wire?. Isn't the tension the same when a wire goes through a pulley? Or am I missunderstanding the message?.
If the tensions were the same on both sides of the pulley, what would be the net torque on the pulley?