A block (mass = 1.9 kg) is hanging from a massless cord that is wrapped around a

Question

A block (mass = 1.9 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.3 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.037 m during the block's descent. Find (a) the angular acceleration of the pulley and (b) the tension in the cord.

Explanation / Answer

Newtons Second says the force on the block is: F = m_b*g-T = m_b*a_y --> T = m_b*g-m_b*a_y Where m_b is the mass of the block. Now we know that ay = dv/dt = r dw/dt = r*alpha where alpha is angular acceleration So T = m_b*g-m_b*r*alpha now the torque on the pulley is tau = r T = I*alpha (I is moment of inertia of pulley). So m_b*g *r - m_b*r^2*alpha = I*alpha ---> alpha = m_b*g*r/(m*r^2+I) now T = I *m_b*g/(m*r^2 +I) from torque equaiton. just plug in the numbers.thank you.

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