Two objects, of masses m_1 = 470.0 g and m_2 = 535.5 g, are connected by a strin

Question

Two objects, of masses m_1 = 470.0 g and m_2 = 535.5 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley a uniform 45.0-g disk with a radius of 3.88 cm. The string does not slip on the pulley. Find the accelerations of the objects. m/s^2 What is the tension in the string between the 470.0-g block and the pulley? N What is the tension in the string between the 535.5-g block and the pulley? N By how much do these tensions differ? N What would your answers be if you neglected the mass of the pulley? acceleration m/s^2 tension N

Explanation / Answer

(a) Let T1 = tension in the rope from to which m1 is attached,

T2 = tension in the rope to which m2 is attached
Let m2 fall by acceleration a and m1 rise by acceleration a
T1 - m1 g = m1 a --------------------(1)
m2 g - T2 = m2 a ----------------(2)
Adding the above two equations,
(T1 - T2) + (m2 - m1)g = (m1 + m2)a
T2 - T1 = (m2 - m1)g - (m1 + m2)a
T2 - T1 = m2(g-a) - m1(g+a) --------------(3)

For the pulley, I = 0.5 mp * R^2 (where mp = mass of the pulley)
Torque = (T2 - T1) * R
Angular acceleration alpha = a/R
Torque = I * alpha
(T2 - T1) * R = 0.5 mp * R^2 * a/R
T2 - T1 = 0.5 mp * a --------------(4)
From equations (3) and (4)
m2(g-a) - m1(g+a) = 0.5 mp * a
m2 * g - m2 * a - m1 * g - m1 * a = 0.5 mp * a
(m2 - m1)g = (m1 + m2 + 0.5 mp)a
a = (m2 - m1)g/(m1 + m2 + 0.5 mp)
a = (0.5355 - 0.4700) * 9.81/(0.4700 + 0.5355 + 0.5 * 0.045)
a = 0.6250 m/s^2
Ans: 0.6250 m/s^2

(b) From equation (1)
T1 = m1 (a + g) = 0.4700(0.6250 + 9.81) = 4.90445 N
Ans: 4.90445 N

From equation (2)
T2 = m2 (g-a) = 0.5355 (9.81 - 0.6250) = 4.9185 N
Ans: 4.9185 N

T2 - T1 = 4.9185 - 4.9044 = 0.0141 N
Ans: 0.0141 N

(c)For pulley mass = 0,

a = g*[1 - 2/(1+m2/m1)] = 0.639m/s2

and T = 2g/(1/m1+1/m2) = 4.911 N

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