Answer:
a) [tex]\omega_f=2.7018\ rad.s^{-1}[/tex]
b) [tex]n=0.068\ rev[/tex]
Explanation:
Given:
diameter of the table, [tex]d=0.75\ m[/tex]
initial angular velocity of the table, [tex]\omega_i=0.25\ rev.s^{-1}=0.5\pi\ rad.s^{-1}[/tex]
angular acceleration of the table, [tex]\alpha=0.9\ rev.s^{-2}=0.9\times2\pi=1.8\pi\ rad.s^{-2}[/tex]
a)
time of observation, [tex]t=0.2\ s[/tex]
Now the angular velocity after the given time:
using equation of motion
[tex]\omega_f=\omega_i+\alpha.t[/tex]
where:
[tex]\omega_f=[/tex] final angular velocity
[tex]\omega_f=0.5\pi\times +1.8\pi\times 0.2[/tex]
[tex]\omega_f=2.7018\ rad.s^{-1}[/tex]
b)
No. of revolutions made in the given time:
[tex]n=N_i.t+\frac{1}{2}\times \alpha.t^2[/tex]
where:
[tex]N_i=[/tex] initial revolution speed
[tex]n=0.25\times0.2+\frac{1}{2}\times 0.9\times 0.2^2[/tex]
[tex]n=0.068\ rev[/tex]
a. The angular velocity of the turntable after 0.200 seconds is 0.43 rev/s.
b. The number of revolutions the turntable spun in this time interval is 0.068 revs.
Given the following data:
a. To compute the angular velocity of the turntable after 0.200 seconds, we would use the first equation of kinematics:
[tex]w_f = w_o + at[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]w_f = 0.250 + 0.900(0.200)\\\\w_f = 0.250 + 0.18[/tex]
Final angular velocity = 0.43 rev/s
b. To find how many revolutions the turntable spun in this time interval, we would use the second equation of kinematics:
[tex]X = w_ot + \frac{1}{2} at^2\\\\X = 0.250(0.2) + \frac{1}{2} (0.9)(0.2)^2\\\\X = 0.05 + 0.018[/tex]
X = 0.068 revs.
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