Question 30: Accelerating Pulley System
Note: The pulley would rotate as the accelerations of blocks on the left and right are different.
System Analysis
Using Newton’s Second Law for the central mass $m_0$:
$$ m_0 g – 2T = \frac{3 m_0 a_2}{2} $$
From the constraints of the system (substitutions implied in the notes):
$$ T = 2 m_0 a_2 $$
Substituting $T$ back into the force equation:
$$ m_0 g – 2(2 m_0 a_2) = \frac{3}{2} m_0 a_2 $$ $$ m_0 g – 4 m_0 a_2 = 1.5 m_0 a_2 $$ $$ m_0 g = 5.5 m_0 a_2 $$Solving for accelerations:
$$ a_2 = \frac{g}{5.5} = \frac{2g}{11} $$ $$ a_1 = 2a_2 = \frac{4g}{11} $$Rotational Kinematics
Assumption: No slipping between the string and the pulley.
Relating linear accelerations of the string to the center of mass acceleration ($a$) and angular acceleration ($\alpha$):
$$ a + \alpha r = a_1 \quad \text{(Left accel)} $$ $$ a – \alpha r = a_2 \quad \text{(Right accel)} $$Subtracting the two equations to isolate $\alpha$:
$$ (a + \alpha r) – (a – \alpha r) = a_1 – a_2 $$ $$ 2\alpha r = a_1 – a_2 $$ $$ \alpha r = \frac{a_1 – a_2}{2} $$Substituting the values of $a_1$ and $a_2$ found earlier:
$$ \alpha r = \frac{\frac{4g}{11} – \frac{2g}{11}}{2} $$ $$ \alpha r = \frac{2g/11}{2} = \frac{g}{11} $$ $$ \alpha = \frac{g}{11r} $$Final Calculation
Given $t = 0.22$ s and $r = 0.1$ m (implied from calculation):
$$ \omega = \alpha t $$ $$ \omega = \frac{g}{11} \times \frac{0.22}{0.1} $$Assuming $g \approx 10 \text{ m/s}^2$:
$$ \omega = \frac{10}{11} \times 2.2 = 2 \text{ rad/s} $$
$$ \omega = 2 \text{ rad/s} $$