Two masses, mA = 30.0 kg and mB = 42.0 kg are connected by a rope that hangs ove

Question

Two masses, mA = 30.0 kg and mB = 42.0 kg are connected by a rope that hangs over a pulley (as in the figure(Figure 1) ). The pulley is a uniform cylinder of radius R = 0.311 m and mass 3.5 kg . Initially, mA is on the ground and mB rests 2.5 m above the ground.

If the system is now released, use conservation of energy to determine the speed of mB just before it strikes the ground. Assume the pulley is frictionless.

Express your answer using three significant figures and include the appropriate units.

Explanation / Answer

Here, the initial total mechanical energy = mBgh.
Final total mechanical energy of the system =0.5(mA +mB) v^2 + mAgh + 0.5Iv^2/r^2.

Where
0.5(mA +mB) v^2 is the kinetic energy of both masses
mAgh is the potential energy of mass mA (The mass mA ascends up a distance h while mB descends a height h.
0.5Iv^2/r^2is the rotational kinetic energy of the pulley
I for cylinder = mpr^2/2 where mp is the mass of the pulley.
0.5Iv^2/r^2 = mv^2

Equating the total energy before and after .

(mB - mA) gh = 0.5(mA +mB +2m) v^2

v^2 = 0.5(mA +mB +2m) / {(mB - mA) gh}

Putting the values in the above expression -

v^2 = 0.5*(30 +42 + 7) / { (42 - 30) *9.8*2.5} = 39.5 / 294 = 0.1343

=> v = 0.366 m/s

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