Respuesta :
Answer:
Explanation:
1 ) tire of radius 0.381 m rotating at 12.2 rpm
12.2 rpm = 12.2 /60 rps
n = .20333 rps
angular speed
= 2πn
= 2 x 3.14 x .20333
= 1.277 rad / s
2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
angular speed = .456 rad/s
3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second
18.7° per second = (18.7 / 180) x 3.14 rad/s
= .326 rad/s
4 )
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2
centripetal acceleration = ω²R
ω is angular velocity and R is radius
4.53 = ω² x .587
ω = 2.78 rad / s
5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
The radius of the circle in which corner is moving
= .123 x √2
=.174 m
angular velocity = linear velocity / radius
.287 / .174
1.649 rad / s
The perfect order is
4 ) > 5> 1 >2>3.
The angular speed of the given objects as follows from the largest to the smallest;
- A square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s.
- A rock on a string being swung in a circle of radius 0.587 m with centripetal acceleration of 4.53 m/s².
- A tire of radius 0.381 m rotating at 12.2 rpm.
- A bowling ball of radius 12.4 cm rotating at 0.456 rad/s
- A top with a diameter of 5.09 cm spinning at 18.7∘ per second
The angular speed of the objects is calculated as follows;
when the radius is 0.381 and rotation of 12.2 rpm;
[tex]\omega = 12.2 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 1.278 \ rad/s[/tex]
The angular speed of the bowling ball = 0.456 rad/s.
A top with a diameter of 5.09 cm spinning at 18.7∘ per second;
[tex]\omega = 18.7 \ ^0/s \ \times \ \frac{2\pi \ rad}{360^0} \\\\\omega = 0.326 \ rad/s[/tex]
A rock on a string being swung in a circle of radius 0.587 m with centripetal acceleration of 4.53 m/s²;
[tex]a_c = \omega^2 r\\\\\omega^2= \frac{a_c}{r} \\\\\omega = \sqrt{\frac{a_c}{r} } \\\\\omega = \sqrt{\frac{4.53}{0.587} } \\\\\omega =2.78 \ rad/s[/tex]
A square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
[tex]radius, \ r = \frac{0.123}{2} = 0.0615\ m\\\\v =\omega r\\\\\omega = \frac{v}{r} \\\\\omega = \frac{0.287}{0.0615} \\\\\omega = 4.67 \ rad/s[/tex]
Thus, we can arrange the angular speed of the given objects as follows from the largest to the smallest;
- A square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s.
- A rock on a string being swung in a circle of radius 0.587 m with centripetal acceleration of 4.53 m/s².
- A tire of radius 0.381 m rotating at 12.2 rpm.
- A bowling ball of radius 12.4 cm rotating at 0.456 rad/s
- A top with a diameter of 5.09 cm spinning at 18.7∘ per second
Learn more here:https://brainly.com/question/14453709
