Engineering Mechanics

Chapter 31 : Kinetics of Motion of Rotation „„„„„ 643

Since the mass is moving upwards with an acceleration (a 2 ), therefore force acting on the
body

= 75 a 2 ...(v)
Equating equations (iv) and (v),
P 2 – 367.5 = 75 a 2 ...(vi)
Now consider the motion of the pulley, which is rotating about its axis due to downward
motion of the 150 kg mass tied to the string. We know that linear acceleration of the 150 kg mass
is equal to the angular acceleration of the pulley.

∴ a 1 =r 1 α = 0.25 α

Similarly a 2 =r 2 α = 0.125 α

and torque, T=P 1 r 1 – P 2 r 2 = P 1 × 0.25 – P 2 × 0.125 ...(vii)

We also know that torque on the pulley,

T=Iα = 0.75 α ...(viii)

Equating equations (vii) and (viii),

0.25 P 1 – 0.125 P 2 = 0.75 α

or P 1 – 0.5 P 2 = 3 α ...(Multiplying by 4)

∴ P 1 = 0.5 P 2 + 3 α ...(ix)

Substituting the value of P 1 in equation (iii),

1039 – (0.5 P 2 + 3 α) = 150 a 1 = 150 × 0.25 α

1039 – 0.5 P 2 – 3 α= 37.5 α

∴ 1039 – 0.5 P 2 = 37.5 α + 3α = 40.5 α

Multiplying both sides by 2

2078 – P 2 = 81 α ...(x)

From equation (vi), we find that

P 2 – 367.5 = 75 a 2 = 75 × 0.125 α = 9.4 α ...(xi)

Adding equations (x) and (xi),

1710.5 = 90.4 α

∴α=

1710.5=18.9 rad/s 2

90.4

Now substituting the value of α in equation (x),

2078 – P 2 = 81 α = 81 × 18.9 = 1531

∴ P 2 = 2078 – 1531 = 547 N Ans.

Again substituting the value of α and P 2 in equation (ix),

P 1 = (0.5 × 547) + (3 × 18.9) = 330 N Ans.

Acceleration of the masses A and B

We know that the acceleration of mass A (i.e., 150 kg),

a 1 =r 1 α = 0.25 × 18.9 = 4.72 m/s^2 Ans.

Similarly a 2 =r 2 α = 0.125 × 18.9 = 2.36 m/s^2 Ans.