Apulley with the radius of 10.0 cm was connected to a motor with a massless
string. the motor turns at the rate of 15000 rad/s. then the motor slow the
turning uniformly to 1 rad/s in 9s. calculate :
(i) the angular acceleration
(ii) the number of turn have been made within that period.
(iii) the length for the string that encircles within that period.
(iv) the tangential acceleration for that string.
(v) the extra time needed for the pulley to stop turning.

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

0.11 m/s and 0.28 m/s should be correct

c. newton's third law of motion