In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m and one of mass 2 hang from either side of the pulley by a light cord. Initially the system is a test with block one on the floor and block two held at the height h above the floor. Block 2 is then released and allowed to fall. Give your answers in terms of m, m2' M, R, h, and g

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14-03-2018
Physics

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In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m and one of mass 2 hang from either side of the pulley by a light cord. Initially the system is a test with block one on the floor and block two held at the height h above the floor. Block 2 is then released and allowed to fall. Give your answers in terms of m, m2' M, R, h, and g

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laurawinter1p43zv5 laurawinter1p43zv5 · Answer 1 · 2018-03-14T15:16:25+01:00

Assuming you are looking for the acceleration a:

1.[tex]m_1a = T_1 -m_1g [/tex]
2.[tex]m_2a = m_2g - T_2[/tex]
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3.[tex]\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R} [/tex]
where [tex]I = \frac{1}{2} mR^2[/tex] and [tex]a = \alpha R[/tex].

Combining the three equations:
[tex]T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M } [/tex]