Two objects, of masses m 1 = 480.0 g and m 2 = 545.7 g, are connected by a strin

Question

Two objects, of masses m1 = 480.0 g and m2 = 545.7 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 48.5-g disk with a radius of 4.16 cm. The string does not slip on the pulley.

(a) Find the accelerations of the objects.
m/s2

(b)1- What is the tension in the string between the 480.0-g block and the pulley? (Round your answer to four decimal places.)
N

2-What is the tension in the string between the 545.7-g block and the pulley? (Round your answer to four decimal places.)
N

3-By how much do these tensions differ? (Round your answer to four decimal places.)
N

(c) What would your answers be if you neglected the mass of the pulley?

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 = m1a --------------------(1)
m2 g - T2 = m2a ----------------(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 = [(0.5457 - 0.480) * 9.81] / (0.480 + 0.5457 + 0.5 * 0.0485)
a = 0.6139 m/s^2

(b) From equation (1)
T1 = m1 (a + g) = 0.480 (0.6139 + 9.81) = 5 N

c) From equation (2)
T2 = m2 (g-a) = 0.5457 (9.81 - 0.6139) = 5.01 N

d) T2 - T1 = 5.01 - 5 = 0.01 N

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