Engineering Mechanics

Chapter 31 : Kinetics of Motion of Rotation „„„„„ 641

Now consider the mass 10 kg, which is going upwards. We know that the forces acting on it
are m 2 g = 10 × 9.8 = 98 newtons (downwards) and P 2 newtons (upwards). As the body is moving
upwards, therefore, resultant force

=P 2 – 98 ...(iv)
Since the mass is moving upwards with an acceleration (a 2 ), therefore force acting on the
body

= 10 a 2 ...(v)
Equating equations (iv) and (v), P 2 – 98 = 10 a 2 ...(vi)
Now consider the motion of the pulley, which is rotating about its axis due to downward
motion of the 30 kg mass tied to the string. We know that the linear acceleration of the body is equal
to the angular acceleration of the pulley.

∴ a 1 =r 1 α = 0.05 α

Similarly a 2 =r 2 α = 0.1 α

and torque, T=P 1 r 1 – P 2 r 2 = P 1 × 0.05 – P 2 × 0.1 ...(vii)

We also know that torque on the pulley, T = Iα = 0.02 α ...(viii)

Equating equations (vii) and (viii),

0.05 P 1 – 0.1 P 2 = 0.02 α

P 1 – 2 P 2 = 0.4 α ...(Multiplying by 20)

P 1 = 0.4 α + 2P 2 ...(ix)

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

294 – (0.4 α + 2P 2 ) = 30 a 1 = 30 × 0.05 α = 1.5 α ...(Q a 1 = r 1 α = 0.05 α)

294 – 0.4 α – 2P 2 = 1.5 α

294 – 2P 2 = 1.5 α + 0.4 α = 1.9 α

Dividing both sides by 2,

147 – P 2 = 0.95 α ...(x)

From equation (vi) we find that

P 2 – 98 = 10 a 2 = 10 × 0.1 α = α ...(xi)

...(Q a 2 = r 2 α = 0.1 α)

Adding equations (x) and (xi),

4 9 = 1.95 α

∴α=^2

49

= 25.1 rad/s

1.95

Ans.

Pulls in the two parts of the string

Substituting the value of α in equation (x),

147 – P 2 = 0.95 α = 0.95 × 25.1 = 23.8

∴ P 2 = 147 – 23.8 = 123.2 N Ans.

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

P 1 = (0.4 × 25.1) + 2 × 123.2 = 256.44 N Ans.