Two objects, of masses m 1 = 460.0 g and m 2 = 504.9 g, are connected by a strin

Question

Two objects, of masses m1 = 460.0 g and m2 = 504.9 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 53.0-g disk with a radius of 3.96 cm. The string does not slip on the pulley.

acceleration m/s2 tension N Two objects, of masses m1 = 460.0 g and m2 = 504.9 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 53.0-g disk with a radius of 3.96 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 460.0-g block and the pulley? (Round your answer to four decimal places.) What is the tension in the string between the 504.9-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

The masses are m1 = 460 g and m2 = 504.9 g

The mass of pulley M = 53 g

The radius of the disk is R = 3.96 cm

(a) Let T1 be the tension in the rope from to which m1 is hanged

            T2 be the tension in the rope to which m2 is hanged
Let a be the acceleration of m2 when the system is released from rest
Then the equations of motion of the objects are

T1 - m1 g = m1 a
m2 g - T2 = m2 a

(T1 - T2) + (m2 - m1)g = (m1 + m2)a
T2 - T1 = (m2 - m1)g - (m1 + m2)a
T2 - T1 = m2(g-a) - m1(g+a)

The moment of inertia of the pulley is

I = 0.5 MR2
The torque acting on the pulley is

? = RT2 - RT1

? = R(T2-T1)

Let the angular acceleration be ?

Then ? = a/R

Therefore the torque

? = I?
R(T2-T1) = (0.5MR2)(a/R)
T2 - T1 = 0.5Ma

Therefore
m2(g-a) - m1(g+a) = 0.5Ma
(m2-m1)g = (m1+m2+0.5M)a
a = (m2-m1)g/(m1+m2+0.5M)
a = (0.5049 - 0.4600)*9.8/(0.460+0.5049+0.5*0.053)
a = 0.4438 m/s2

(b)

The tension in the rope from to which m1 is hanged

T1 = m1 (a + g)

      = 0.460 (0.4438 + 9.8)

      = 4.7121 N

c)

The tension in the rope to which m2 is hanged
T2 = m2 (g-a)

      = 0.5049(9.8 - 0.4438)

      = 4.7239 N

d)

The difference in the tensions is

?T =  T2 - T1

      = 4.7239 - 4.7121

      = 0.0118 N

e)

If the mass of the pulley is neglected,

Then the tension in the string will be same on both sides

T1 = T2 = T

Then

(m2 - m1)g = (m1 + m2)a

a = (m2 - m1)g/(m1 + m2)

a = 0.456 m/s2

The tension in the string T = m1(a+g)

T = 4.71776 N

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