Question

A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the first 3.70 rev and (b) the next 3.70 rev?

Answer 1

Answer:

(a) t = 5.66 s

(b) t = 8 s

Explanation:

(a)

Here we will use 2nd equation of motion for angular motion:

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (23.25 rad)(2)/(1.45 rad/s²)

t = √(32.06 s²)

t = 5.66 s

(b)For next 3.7 rev

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (46.5 rad)(2)/(1.45 rad/s²)

t = √(64.13 s²)

t = 8 s

Answer 2

The time taken in each case is  2.55 s

Let us recall that the equations of circular motion are almost like those of linear motion;

α = ω2 - ω1/t

α = angular acceleration

ω2 = final angular velocity

ω1 = initial angular velocity

t = time taken

a)  1.45 rad/s2 =3.70 rev/s - 0 rev/s/t

t = 3.70 rev/s/1.45 rad/s2

t = 2.55 s

b) For, ) the next 3.70 rev:

1.45 rad/s2 =3.70 rev/s - 0 rev/s/t

t = 3.70 rev/s/1.45 rad/s2

t = 2.55 s

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