Respuesta :
Answer:
a. 70 rad/s²
b. 5000 rev
Explanation:
As we know,
[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]
then,
[tex]\omega=2100 \ rad/s[/tex]
a...
⇒ [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2100}{30}[/tex]
⇒ [tex]=70 \ rad/s^2[/tex]
b...
⇒ [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\times 70\times (30)^2[/tex]
[tex]=31500 \ rad[/tex]
[tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]
[tex]=5000 \ rev[/tex]
(a) The average angular acceleration will be 70 rad/s².
(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
What is angular acceleration?
Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².
The given data in the problem is;
n is the revolution of centrifugal rotor = 20000 rpm
t is the time interval= the 30s
(α) is the Angular acceleration=?
n₁ is the revolution when the acceleration is constant =?
(a) The average angular acceleration will be 70 rad/s².
The value of the angular velocity is given by
[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]
The formula for angular acceleration is guven by;
[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]
Hence the average angular acceleration will be 70 rad/s².
(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.
[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]
As we know that the angular velocity is given by
[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]
The relation of angular velocity and revolution will be
[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]
Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
To learn more about angular acceleration refer to the link ;
https://brainly.com/question/408236
