The two blocks in the figure(Figure 1) are connected by a massless rope that pas
ID: 2194521 • Letter: T
Question
The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 17in diameter and has a mass of 2.2 . As the pulley turns, friction at the axle exerts a torque of magnitude 0.47 If the blocks are released from rest,
how long does it take the 4.0 block to reach the floor?
Express your answer to two significant figures and include the appropriate units.
The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 17in diameter and has a mass of 2.2 . As the pulley turns, friction at the axle exerts a torque of magnitude 0.47 If the blocks are released from rest, how long does it take the 4.0 block to reach the floor? Express your answer to two significant figures and include the appropriate units.Explanation / Answer
Let: m1 be the left hand mass, T1 be the left hand tension, m2 be the right hand mass, T2 be the right hand tension, g be the acceleration due to gravity, a be the acceleration of the masses, alpha be the angular acceleration of the pulley, r be the radius of the pulley (assumed solid), F be the friction couple, h be the descent for m1, t be the time taken. For mass m1: m1 g - T1 = m1 a T1 = m1(g - a) For mass m2: T2 - m2 g = m2 a T2 = m2(g + a) For the pulley: (T1 - T2)r - F = I alpha = (m3 r^2 / 2)(a / r) T1 - T2 = F / r + m3 a / 2 Eliminating T1 and T2: m1(g - a) - m2(g + a) = F / r + m3 a / 2 (m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r (m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2) h = at^2 / 2 t = sqrt(2h / a) t= sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ] } 17inch= 17x 0.0254= 0.43m Taking the pulley mass to be 3 kg (the question gives 3.0 cm): t = sqrt{ 1( 2 * 4 + 2 * 2 + 2.2) / [ 0.47 / 0.21 + (4 - 2)9.81 ] } = 0.649 sec.
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