Engineering Mechanics

Chapter 21 : Motion of Rotation „„„„„ 451

Now consider the motion of the bridge in the next 60 sec. In this case, constant angular
velocity of (40 α 1 ) as obtained in equation (i) and time (t 2 ) = 60 sec. Therefore angular displacement
during 60 sec,
θ 2 = 40 α 1 × t 2 = 40 α 1 × 60 = 2400 α 1 rad ...(iii)
Now consider the motion of the bridge in the last 20 sec. In this case, initial angular velocity
(ω 0 ) = 40 α 1 as obtained in equation (i) ; Final angular velocity (ω) = 0 and time (t 3 ) = 20 sec. We
know that final angular velocity of the bridge (ω)
0=ω 0 – αt 3 = 40 α 1 – α 2 × 20 or α 2 = 2α 1
...(Minus sign due to retardation)

and angular displacement, 303 23221 1

11

(40 20) – 2 (20)

22

θ=ω − α =ttα× ×α

= 400 α 1 ...(iv)

...(Minus sign due to retardation)

We also know that total angular displacement of the bridge (θ)

0.5π = θ 1 + θ 2 + θ 3 = 800 α 1 + 2400 α 1 + 400 α 1 = 3600 α 1

∴ 1 –3^2

0.5

0.436 10 rad/sec

3600

π

α= = × Ans.

(ii) Maximum angular velocity of the bridge

We know that maximum angular velocity of the bridge,
ω = 40 α 1 = 40 × 0.436 × 10–3 = 17.44 × 10–3 rad/s Ans.
(iii) Angular retardation of the bridge

We also know that angular retardation of the bridge,
α 2 = 2α 1 = 2 × ( 0.436 × 10–3 ) = 0.872 × 10–3 rad/sec^2 Ans.
Example 21.8. A flywheel rotates with a constant retardation due to braking. From t = 0 to
t = 10 seconds, it made 300 revolutions. At time t = 7.5 sec, its angular velocity was 40 π rad/sec.
Determine (i) value of constant ratardation ; (ii) total time taken to come to rest and (iii) total
revolutions made till it comes to rest.

Solution. Given : Time interval (t) = 10 – 0 = 10 sec ; Angular displacement (θ) = 300
rev = 2π × 300 = 600 π rad ; when time (t) = 7.5 sec and angular velocity (ω) = 40 π rad/sec

(i) Value of constant retardation
Let α= Constant retardation, and
ω 0 = Initial angular velocity.
First of all, consider motion of the flywheel from t = 0 to t = 10 seconds. We know that total
angular displacement (θ)

()

22

00

11

600 – 10 – (10)

22

tt
⎛⎞
π=ω α = ω × ⎜⎟α
⎝⎠
= 10 ω 0 – 50α ...(i)
...(Minus sign due to retardation)
Now consider motion of the flywheel from t = 0 to t = 7.5 seconds. We know that final angular
velocity (ω)
40 π= ω 0 – αt = ω 0 – α × 7.5 ...(ii)
...(Minus sign due to retardation)
0 –40 rad/sec 2
7.5

ωπ

α= ...(iii)