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Two blocks, which can be modeled as point masses, are connected by a massless string which passes through a hole in a frictionless table. A tube extends out of the hole in the table so that the portion of the string between the hole and M1 remains parallel to the top of the table. The blocks have masses M1 = 1.9 kg and M2 = 2.8 kg. Block 1 is a distance r = 0.95 m from the center of the frictionless surface. Block 2 hangs vertically underneath. find the speed of m1 assume m2 does not move relative to the table.

Answer :

Onyebuchinnajigo

The speed of the block m1 on the frictionless table is 1.34 m/s.

The given parameters;

  • mass of the first block, m1 = 1.9 kg
  • mass of the second block, m2 = 2.8 kg
  • distance of block m1, R = 0.95 m

The net torque on both blocks is calculated as;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_2R- T_1 R_1 = I \alpha \\\\[/tex]

where;

  • T₁ is the tension on first block
  • I is the moment of inertia of point mass
  • α is the angular acceleration

[tex]T_1 = m_1 g + m_1 a\\\\T_2 = m_2 g - m_2 a[/tex]

The acceleration of both blocks is calculated as follows;

[tex]R(T_ 2- T_1) = MR^2 \times (\frac{a}{R} )\\\\R(T_2 -T_1) = MRa\\\\T_2 - T_1 = Ma\\\\(m_2g - m_2 a) - (m_1 g + m_1 a) = Ma\\\\m_2 g - m_1 g - m_2 a - m_1 a = Ma\\\\g(m_2 - m_1) = Ma + m_2a+ m_1a\\\\g(m_2 - m_1) = a(M+ m_2 + m_1)\\\\where;\\\\M \ is \ mass \ of \ string = 0 \\\\g(m_2 - m_1) = a (0+ m_2 + m_1)\\\\g(m_2 - m_1) = a(m_1 + m_2)\\\\a = \frac{g(m_2 - m_1)}{m_1 + m_2} \\\\a = 1.88 \ m/s^2[/tex]

The speed of the block m1 is calculated as follows;

[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{a r} \\\\v = \sqrt{1.88 \times 0.95} \\\\v = 1.34 \ m/s[/tex]

Thus, the speed of the block m1 on the frictionless table is 1.34 m/s.

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