Question

Suppose you have a system of two masses strung over a pulley. Mass 1 (6.7 kg) hangs on the right side of the pulley suspended over the ground at height 1.2 m. Mass 2 (2.5 kg) hangs over the left side of the pulley and rests on the ground. The pulley is a uniform disk of mass m and radius 11.3 cm. When the system is released, Mass 1 moves down and Mass 2 moves up, such that Mass 1 strikes the ground with speed 1.1 m/s. This is called an Atwood Machine. Calculate the mass of the pulley in kg.

Answer 1

we are given initial height of mass m1 and its speed on impact. so, we can use kinematics to find the acceleration

v² - u² = 2ah

1.1² - 0 = 2 *a * 1.2

a = 0.50416 m/s²

Now, we know for the given atwood system, the acceleration is given as

a = ( m1 - m2)g / m1 + m2 + 1/2m_p ( let me know if you need the derivation of this)

where m1 and m2 are masses of blocks and m_p is mass of pulley

so,

0.50416 = ( 6.7 - 2.5) * 9.8 / 6.7 + 2.5 + 1/2m_p

0.50416 = 41.16 / 9.2 + 1/2m_p

9.2 + 1/2m_p = 41.16 / 0.50416

9.2 + 1/2m_p = 81.64

so,

m_p = 144.88 Kg

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Here is the derivation of the bold equation

torque = I

T1 * R - T2 * 2 = 1/2 * m_p * r² *

T1 * R - T2 * R = 1/2 * m_p * R² * a / R

T1 * R - T2 * R = 1/2 * m_p * R * a

(m1g - m1a ) - ( m2g - m2a) = 1/2 * m_p  * a

m1g - m2g = (1/2m_p + m1 + m2 ) a

a = m1g - m2g / 1/2m_p + m1 + m2