Two objects, of masses m1 = 530.0 g and m2 = 555.9 g, are connected by a string

Question

Two objects, of masses m1 = 530.0 g and m2 = 555.9 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform 55.0-g disk with a radius of 3.84 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 530.0-g block and the pulley? (Round your answer to four decimal places.) What is the tension in the string between the 555.9-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? Acceleration? Tension?

Explanation / Answer

After the masses have each moved a distance h, the changes in gravitational potential energy are

U1 =-m1gh

U2 =m2gh

Since mechanical energy is conserved,

U+K=0

The mechanical energy of the system includes the kineticenergies of three objects: the two masses and the pulley.All start with zero kinetic energy, so

K=(1/2)(m1+m2)V2+(1/2)I2

-m1gh+m2gh+(1/2)(m1+m2)V2+(1/2)I2=0

m1gh =m2gh+(1/2)(m1+m2)V2+(1/2)I2

I=(1/2)mr2 ,=V/r

m1gh =m2gh+(1/2)(m1+m2)V2+(1/2)(1/2)mr2(V/r)2

V=[(2gh(m2-m1)/(m1+m2+(1/2)m)]

1)acceleration a=V2/2h

a=(m2-m1)g/(m1+m2+(1/2)m) =(555.9-530)*9.81/(555.9+530+(1/2)*55) =0.228m/s or 22.8cm/s

2)T1 =m1(g+a) =530*10-3(9.81+0.228) =5.3201N

T2 =m2(g-a) =555.9*10-3(9.81-0.228) =5.3266N

T =T2-T1 =5.3266-5.3201 =0.0065 N or 6.5mN

C)if mass of pulley is neglected means i.e m=0

a=(m2-m1)g/(m1+m2) =(555.9-530)*9.81/(555.9+530) =0.234m/s2 or 23.4cm/s2

T=T1 =T2 =m1(g+a) =530*10-3(9.81+0.234) =5.32 N

so T =0

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