Respuesta :
First, let's convert the initial angular speed from rpm (rev/min) into rad/s, keeping in mind that [tex]1 rev = 2 \pi rad[/tex] and 1 min=60 s:
[tex]\omega_i = 33 \frac{rev}{min}=33\frac{rev}{min} \cdot \frac{2 \pi rad}{60 s}=3.45 rad/s [/tex]
Now we can find the angular acceleration of the lp, keeping in mind that the final speed is zero:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}= \frac{0-3.45 rad/s}{5.5 s}=-0.63 rad/s^2 [/tex]
and the acceleration is negative because the LP is decelerating.
Now we can find the angle covered by the LP from the beginning to the end of its motion:
[tex]\theta (t)= \omega_i t + \frac{1}{2}\alpha t^2 = (3.45 rad/s)(5.5 s)+ \frac{1}{2}(-0.63 rad/s^2)(5.5 s)^2=[/tex]
[tex]=9.45 rad [/tex]
And finally, we can convert it into number of revolutions:
[tex]1 rev : 2 \pi rad = x: 9.55 rad[/tex]
[tex]x= \frac{9.55}{2 \pi}= 1.52 rev[/tex]
[tex]\omega_i = 33 \frac{rev}{min}=33\frac{rev}{min} \cdot \frac{2 \pi rad}{60 s}=3.45 rad/s [/tex]
Now we can find the angular acceleration of the lp, keeping in mind that the final speed is zero:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}= \frac{0-3.45 rad/s}{5.5 s}=-0.63 rad/s^2 [/tex]
and the acceleration is negative because the LP is decelerating.
Now we can find the angle covered by the LP from the beginning to the end of its motion:
[tex]\theta (t)= \omega_i t + \frac{1}{2}\alpha t^2 = (3.45 rad/s)(5.5 s)+ \frac{1}{2}(-0.63 rad/s^2)(5.5 s)^2=[/tex]
[tex]=9.45 rad [/tex]
And finally, we can convert it into number of revolutions:
[tex]1 rev : 2 \pi rad = x: 9.55 rad[/tex]
[tex]x= \frac{9.55}{2 \pi}= 1.52 rev[/tex]
No of revolution is defined as the no of loops taken from the starting to the end from the starting point . The number of rotations did it turn while stop will be 1.52.
What is no revolution?
No of revolution is defined as the no of loops taken from the starting to the end from the starting point.
The given data in the problem is;
n is the rpm of the old Ip =33.3 rpm
t is time interval taken to stop= 5.5 sec
N is the number of rotations did it turn while stopping=?
The angular speed is found by the formula as;
[tex]\rm \omega = \frac{2\pi N}{60} \\\\ \rm \omega = \frac{2\times 3.14 \times 33.3}{60} \\\\ \rm \omega = 3.45 \ rad/sec[/tex]
The rate of change of angular velocity is known as the angular acceleration;
[tex]\rm \alpha = \frac{\omega_f - \omega_i}{t} \\\\ \rm \alpha = \frac{ 0-3.45}{5.5} \\\\ \rm \alpha =-0.63\ rad /sec^2[/tex]
The equation for the newtons law is given as
[tex]\rm \theta(t)= \omega_0+\frac{1}{2}\alpha t^2 \\\\ \rm \theta(t)= 3.45+\frac{1}{2}\times (-0.63) (5.5)^2 \\\\ \rm \theta(t)= 9.45\ rad[/tex]
For one rotation the number of revolutions is n then for the N no of revolutions;
[tex]\rm \frac{1}{2\pi} = \frac{N}{\theta} \\\\ N=\frac{\theta}{2\pi} \\\\ \rm N=\frac{9.45}{2\times 3.14} \\\\ \rm N= 1.52\ revolution[/tex]
Hence the number of rotations did it turn while stop will be 1.52.
To learn more about the revolution refer to the link;
https://brainly.com/question/1291142
