Two objects, of masses m 1 = 495.0 g and m 2 = 525.3 g, are connected by a strin

Question

Two objects, of masses m1 = 495.0 g and m2 = 525.3 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 46.5-g disk with a radius of 4.16 cm. The string does not slip on the pulley.

acceleration m/s2 tension N Two objects, of masses m1 = 495.0 g and m2 = 525.3 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 46.5-g disk with a radius of 4.16 cm. The string does not slip on the pulley. (a) Find the accelerations of the objects. (b) What is the tension in the string between the 495.0-g block and the pulley? (Round your answer to four decimal places.) What is the tension in the string between the 525.3-g block and the pulley? (Round your answer to four decimal places.) By how much do these tensions differ? (Round your answer to four decimal places.) (c) What would your answers be if you neglected the mass of the pulley?

Explanation / Answer

I am giving the complete solution.

(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 = (0.5253 - 0.495) * 9.81/(0.495 + 0.5253 + 0.5 * 0.0465)
a = 0.2848 m/s^2
Ans: 0.2848 m/s^2

(b) From equation (1)
T1 = m1 (a + g) = 0.495 (0.2848 + 9.81) = 4.9969 N
Ans: 4.9969 N

c) From equation (2)
T2 = m2 (g-a) = 0.5253 (9.81 - 0.2848) = 5.00 N
Ans: 5.0035 N

d) T2 - T1 = 5.0038 - 4.9969 = 0.0069 N
Ans: 0.0069 N
_________________.

Note:
I have taken g = 9.81 m/s^2
The values are given in grams. I converted into kg

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