Respuesta :
Answer
given,
time period, t = 0.145 s
angular velocity, ω = 2.85 x 10⁴ rev/ min
[tex]\omega =2.85\times \dfrac{2 \pi}{60}[/tex]
[tex]\omega =2251.47\ rad/s[/tex]
using rotational motion equation
[tex]\omega =\omega _{o}+\alpha t[/tex]
Since it starts from rest ,initial angular velocity ωo=0
final angular velocity
[tex]\alpha=\dfrac{\omega}{t}[/tex]
[tex]\alpha=\dfrac{2251.47}{0.145}[/tex]
α = 15,527 rad/s²
b)
again using equation of rotational motion
[tex]\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}[/tex]
[tex]\theta =\frac{1}{2}\times 15527\times 0.145^{2}[/tex]
θ = 163.23 rad
a. The drill's angular acceleration (in [tex]rad/s^2[/tex]) is equal to 20582.76 [tex]rad/s^2[/tex]
b. The angle (in radians) through which the drill rotates during this period is 216.13 rads.
Given the following data:
- Time period = 0.145 seconds
- Angular velocity = [tex]2.85 \times 10^4[/tex] rev/min
a. To determine the drill's angular acceleration (in [tex]rad/s^2[/tex]):
First of all, we would convert the value of the angular velocity in rev/min to rad/secs.
Conversion:
1 rev/min = 0.1047 rad/secs.
[tex]2.85 \times 10^4[/tex] rev/min = 2984.5 rad/secs.
Mathematically, angular acceleration is given by the formula:
[tex]\alpha = \frac{\omega}{t}[/tex]
Where:
- [tex]\omega[/tex] is the angular velocity of an object.
- t is the time in seconds.
Substituting the parameters into the formula, we have;
[tex]\alpha =\frac{2984.5}{0.145} \\\\\alpha =20582.76\; rad/s^2[/tex]
b. To determine the angle (in radians) through which the drill rotates during this period:
[tex]\theta = \omega_o t + \frac{1}{2} \alpha t^2\\\\\theta = 0(0.145) + \frac{1}{2} \times 20582.76 \times 0.145^2\\\\\theta = 10291.37 \times 0.0210\\\\\theta = 216.13 \;rads[/tex]
Angle = 216.13 rads.
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