Question

Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall.
The left box has a mass m1 = 4.3 kg and the right box has a mass m2 = 2.1 kg. The pulley has mass m3 = 2.5 kg and radius R = 0.15 m. 1)

1) What is the linear acceleration of the left box? (up is the positive direction) m/s2

2) What is the angular acceleration of the pulley? (let counter-clockwise be positive) rad/s2

3) What is the tension in the string between the left mass and the pulley? N

4) What is the tension in the string between the right mass and the pulley? N

5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s? m

6) What is the magnitude of the velocity of the boxes after the time 1.51 s? m/s

7) What is the magnitude of the final angular speed of the pulley? rad/s

Answer 1

(1) The linear acceleration of the left box is 3.0 m/s².

(2) The angular acceleration of the pulley is 20 rad/s².

(3) The tension in the string between the left mass and the pulley is 29.24 N.

(4) The tension in the string between the right mass and the pulley is 26.88 N.

(4) The distance travelled by each mass for the given time is 3.42 m.

(5) The magnitude of the block's velocity after the given time is 4.53 m/s.

(6) The magnitude of the final angular velocity is 30.2 rad/s.

The given parameters;

• left box, m₁ = 4.3 kg
• right box, m₂ = 2.1 kg
• mass of pulley, m₃ = 2.5 kg
• radius of the pulley, R = 0.15 m

The linear acceleration of the masses is calculated as follows;

$\tau _{net} = I \alpha \\\\T_1R - T_2R = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\(T_1 - T_2)R = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{M}{2} (T_1 - T_2 )\\\\a = \frac{m_3}{2} (T_1 - T_2)\\\\a =\frac{m_3}{2} [m_1g - m_1a - (m_2g + m_2a)]\\\\a = \frac{m_3}{2} (m_1g - m_2 g) - \frac{m_3}{2} (m_1a + m_2a)\\\\a + \frac{m_3}{2} (m_1a + m_2a)\ = \frac{m_3}{2} (m_1g - m_2 g)\\\\a (1 + \frac{m_3}{2} (m_1 + m_2))= \frac{m_3g}{2} (m_1 - m_2 )$

$a = \frac{\frac{m_3g}{2} (m_1 - m_2 )}{1 + \frac{m_3}{2}(m_1 + m_2) } \\\\a = \frac{\frac{2.5\times 9.8}{2} (4.3 - 2.1 )}{1 + \frac{2.5}{2}(4.3 + 2.1) }\\\\a = 3.0 \ m/s^2$

The angular acceleration of the pulley is calculated as;

$\alpha = \frac{a}{R} = \frac{3}{0.15} = 20 \ rad/s^2$

The tension in the string between the left mass and the pulley;

$T_1 = m_1g - m_1a\\\\T_1 = m_1(g -a)\\\\T_1 = 4.3(9.8 - 3)\\\\T_1 = 29.24 \ N$

The tension in the string between the right mass and the pulley;

$T_2 = m_2a + m_2 g\\\\T_2 = m_2(a + g)\\\\T_2 = 2.1(3 + 9.8)\\\\T_2 = 26.88 \ N$

The distance travelled by each mass for the given time is calculated as;

$d = v_0 t + \frac{1}{2} at^2\\\\d = 0 + \frac{1}{2} \times 3.0 \times (1.51)^2\\\\d = 3.42 \ m$

The magnitude of the block's velocity after the given time is calculated as;

$v = at\\\\v = 3 \times 1.51\\\\v = 4.53 \ m/s$

The magnitude of the final angular velocity is calculated as follows;

$\omega _f = \frac{v}{r} \\\\\omega _f = \frac{4.53}{0.15} = 30.2 \ rad/s$

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