Question

A mass m1 is connected by a light string that passes over a pulley of mass M to a mass m2 as shown in the figure. Both masses move vertically and there is no slippage between the string and the pulley. The pulley has a radius of 20.0 cm and a moment of inertia of ½ MR2. If m1 is 3.00 kg, m2 is 6.00 kg and M is 4.00 kg, then what is the acceleration of the masses?

Answer 1

Answer 2

Given Data:

• $m_1 = 3.00 \, \text{kg}$

• $m_2 = 6.00 \, \text{kg}$

• $M = 4.00 \, \text{kg}$ (mass of the pulley)

• Radius of the pulley: $R = 0.20 \, \text{m}$

• Moment of inertia of the pulley:

$$I = \frac{1}{2} M R^2 = \frac{1}{2} \times 4.00 \times (0.20)^2 = \frac{1}{2} \times 4.00 \times 0.04 = 0.08 \, \text{kg m}^2$$

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Step 1: Define the equations of motion.

For $m_1$:

$$T_1 - m_1 g = m_1 a$$$$T_1 = m_1 g + m_1 a$$

For $m_2$:

$$m_2 g - T_2 = m_2 a$$$$T_2 = m_2 g - m_2 a$$

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Step 2: Apply torque equation for the pulley.

Net torque on the pulley:

$$\tau = R(T_2 - T_1) = I \alpha$$

Since $\alpha = \frac{a}{R}$, we get:

$$R(T_2 - T_1) = I \frac{a}{R}$$

Substitute $I = 0.08 \, \text{kg m}^2$:

$$R(T_2 - T_1) = 0.08 \frac{a}{R}$$$$(T_2 - T_1) = \frac{0.08 a}{R^2}$$$$(T_2 - T_1) = \frac{0.08 a}{(0.20)^2} = \frac{0.08 a}{0.04} = 2a$$

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Step 3: Set up the system of equations.

$$m_2 g - m_2 a - m_1 g - m_1 a = 2a$$$$m_2 g - m_1 g = m_1 a + m_2 a + 2a$$$$m_2 g - m_1 g = (m_1 + m_2 + 2)a$$$$(6.00 \times 9.8) - (3.00 \times 9.8) = (3.00 + 6.00 + 2) a$$$$58.8 - 29.4 = 11a$$$$29.4 = 11a$$$$a \approx \frac{29.4}{11} \approx 2.67 \, \text{m/s}^2$$

So, The acceleration of the masses is approximately 2.7 m/s².