Question

A disk rotates about its central axis starting from rest

andaccelerates with constant angular acceleration. At onetime it is
rotating at 10rev/s; 60 rrevolutions later, its angularspeed is
15rev/s. Calculate a) the angular acceleration b)thetime required
to complete the 60 revolutions, c) the time requiredto reach the
10rev/s angular speed, and d) the number ofrevolutions from rest
until the time the disk reaches the 10rev/sangular speed.

Answer 1

a α = 6.54 rad / s², b t = 4.81 s , c t = 9.62 s , d θ = 48.12 rev

Step-by-step explanation:

We can solve this exercise using angular kinematics

Let's reduce the magnitudes to the SI system

w₀ = 10 rev / s (2π rad / 1 rev) = 20π rad / s

`w_(f)` = 15 rev / s = 30π rad / s

θ = 60 rev (2π rad / 1rev) = 120π rad

a)
`w_(f)`² = w₀² + 2 α θ

α = (wf² - w₀²) / 2 θ

α = (30²pi² - 20² pi²) / 2 120 pi

α = 2.08π rad / s²

α = 6.54 rad / s²

b)
`w_(f)` = w₀ + α t

t = (
`w_(f)` - w₀) / α

t = (30π -20ππ) /2.08π

t = 4.81 s

c) the time to reach the speed of wf = 20pi rad / s from rest w₀ = 0

t = (20π -0) /2.08π

t = 9.62 s

d) the revolutions to reach this speed

θ = w₀ t + ½ alf t²

θ = ½ 2.08 pi 9.62²

θ = 302.37 rad

Let's reduce to revolutions

θ = 302.37 rad (1 rev / 2pi rad)

θ = 48.12 rev

Answer 2

To solve this problem we will apply the kinematic equations of angular motion. For this purpose, we will begin by calculating the angular acceleration from the number of revolutions provided against the change in speed. The time will then be calculated, based on the difference in angular velocities and angular acceleration. Finally, the number of revolutions will be obtained from the relationship between time and acceleration.

PART A ) Angular acceleration

`\omega^2 - \omega_0^2 = 2\alpha \theta`

Solving for
`\alpha`:

`\alpha= ((15^2 - 10^2))/((2*60))`

`\alpha = 1.0416 rev/s^2`

PART B) Time required to complete the 60 revolutions

`\omega= \omega_0 + \alpha t`

`t = (\omega-\omega_0)/(\alpha)`

`t =(15-10)/(1.0416)`

`t = 4.8s`

PART C) Time required to complete 10 rev/s from rest

`\omega= \omega_0 + \alpha t`

`t = (\omega-\omega_0)/(\alpha)`

`t =(10-9)/(1.0416)`

`t = 0.96s`

PART D) The number of revolutions from rest until the time the disk reaches the 10 rev/s

`\theta = (1)/(2) \alpha t ^2`

`\theta = (1)/(2) (1.0416)(0.96)^2`

`\theta = 0.478 rev`