PLEASE GIVE COMPLETE EXPLANATION FOR EACH STEP TAKEN TO SOLVE THE PROBLEM! PART

Question

PLEASE GIVE COMPLETE EXPLANATION FOR EACH STEP TAKEN TO SOLVE THE PROBLEM!

PART A answer must equal: -1.96 m/s^2

PART B answere must equal : - 0.196 m/s^2

A 60 kg mass and a 40 kg mass arc suspended on a pulley as shown and held stationary. When they arc released, what will be the acceleration of the masses? [Assume that the moment of inertia of the pulley can be ignored.] What will be the acceleration of the masses if the moment of inertia of the pulley is given by I =(1/2)mpr2 where mP is the mass of the pulley (= 2000 kg), r is its radius.

Explanation / Answer

let acceleration = a and tension in string = T

so

T-60g = - 60a---------(1)

and T - 40g = 40a

T = 40 g+40a

Substitute T value in equation (1)
40g+40a -60g = -60a

-20g + 100a = 0

20g = 100 a

acceleration a = g/5 = 9.8/5 = 1.96 m/s^2

b)
   Torque = I (alpha)

60 g * r - 40 g * r = I (alpha)

we know ( a = r *alpha)

20 g r = I(a/r)

20*9.8* r^2 /(0.5mr^2) = a

196/(0.5*2000) = a

accelration a = 196/1000 = 0.196 m/s^2

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