Respuesta :
Answer:
Angular acceleration = -1.57 Rad/s^2
Number of revolutions = 800 revolutions
Explanation:
Mass = 1.15kg
diameter of the grinding wheel = 22.0cm
angular speed = 20 rev/s
1 Rev/s = 6.283 rad/s
20 Rev/s = 20 * 6.283 = 125.664 rad/s
Angular acceleration = Angular speed/ time taken for for the wheel rotation
Angular accleration = 125.664/80
Angular acceleration = -1.57 rad/s^2 (since the wheel is rotating counterclockwise)
number of revs = 0.5∝t²
number of revs = 0.5 * (1.57/2π) * 80²
number of revs = 800 revolutions
Complete Question:
A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest.
(a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive?
(b) How many revolutions did the wheel make during the time it was coming to rest?
Answer:
(a) the angular acceleration is -1.571 rad/s²
(b) the number of revolutions made by the wheel is 800 revolutions.
Explanation:
Given;
mass of the grinding wheel = 1.15 kg
diameter of the wheel = 22.0 cm
number of revolution = 20
time taken to come to rest = 80.0 s
Part (a) the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive.
[tex]\omega _f = \omega_i + \alpha t[/tex]
where;
ωf is the final angular speed
ωi is the initial angular speed
α is the angular acceleration
[tex]\omega _f = \omega_i + \alpha t\\\\0 = (2\pi *20) + 80 \alpha\\\\ 80 \alpha = -40\pi \\\\\alpha = \frac{-40\pi}{80} = -1.571 \frac{rad}{s^2}[/tex]
Part (b) number of revolutions the wheel made during the time it was coming to rest
[tex]\theta = \omega_i t+\frac{1}{2} \alpha t^2\\\\ \theta = \frac{1}{2}(\frac{1.571}{2\pi } ) (80)^2 = 800, revolutions[/tex]
