Respuesta :
a) 2.22 rad/s
b) 6.66 m/s
c) 0.56 rad/s, 3.36 m/s
Explanation:
a)
The angular velocity of an object in rotation is equal to the rate of change of the angular displacement:
[tex]\omega=\frac{\Delta \theta}{\Delta \theta}[/tex]
where
[tex]\Delta \theta[/tex] is the angular displacement
[tex]\Delta t[/tex] is the time elapsed
First of all, we calculate the full area of the circle. The radius is the length of the sling:
r = 3 m
So the total area is
[tex]A=\pi r^2 = \pi(3)^2=28.3 m^2[/tex]
This area corresponds to an angle of [tex]2\pi[/tex] radians.
Here the area swept out by the sling is
[tex]A'=50 m^2[/tex]
So the angle corresponding to this area is
[tex]\Delta \theta = 2\pi \frac{A'}{A}=2\pi \frac{50}{28.3}=11.1 rad[/tex]
While the time elapsed is
[tex]\Delta t=5 s[/tex]
Therefore, the angular velocity is:
[tex]\omega=\frac{11.1}{5}=2.22 rad/s[/tex]
b)
For an object in rotational motion, the relationship between angular velocity and linear speed is:
[tex]v=\omega r[/tex]
where
v is the linear speed
[tex]\omega[/tex] is the angular velocity
r is the radius
In this problem:
[tex]\omega=2.22 rad/s[/tex] is the angular velocity
r = 3 m is the radius (length of the sling)
So, the speed is:
[tex]v=(2.22)(3)=6.66 m/s[/tex]
c)
In this case, the length of the sling is doubled, so
r = 6 m
First of all, we can re-calculate the area of the full circle:
[tex]A=\pi r^2=\pi (6)^2=113.0 m^2[/tex]
Therefore, the angular displacement of the sling this time is:
[tex]\Delta \theta=2\pi \frac{A'}{A}=2\pi \frac{50}{113}=2.78 rad[/tex]
And so, the angular velocity this time is
[tex]\omega=\frac{\Delta \theta}{\Delta t}=\frac{2.78}{5}=0.56 rad/s[/tex]
The linear speed instead will be given by
[tex]v=\omega r[/tex]
And by substituting the new values, we get:
[tex]v=(0.56)(6)=3.36 m/s[/tex]
(a) The angular velocity is [tex]2.22\, rad/s[/tex].
(b) The linear speed is [tex]6.66\,m/s[/tex].
(c) When the radius is doubled, the angular velocity becomes [tex]0.554 \, rad/s[/tex] and linear velocity becomes [tex]3.324\,m/s[/tex].
The answer is explained as follows.
(a) Given that [tex]r = 3\,m[/tex], [tex]A = 50\,m^2[/tex] and [tex]t = 5 \,s[/tex].
We know that angular velocity is given by,
- [tex]\omega = \frac{\theta}{t}[/tex]
But the area of a sector of a circle is given by,
- [tex]A = \frac{\theta}{2\pi \,rad} \times \pi r^2[/tex]
Now, substituting the values of the given area, and given radius, we get [tex]\theta[/tex], i.e.; the angular displacement.
[tex]50\,m^2 = \frac{\theta}{2\pi\, rad} \times \pi \times (3\,m)^2[/tex]
- [tex]\implies \theta = \frac{50\,m^2 \times 2\pi \, rad }{\pi \times9\,m^2} = 11.1 \,rad[/tex]
Therefore, substituting the values of [tex]\theta[/tex] and [tex]t[/tex] in the equation for angular velocity, we get;
- [tex]\omega = \frac{11.11\,rad}{5\,s}=2.22\,rad/s[/tex].
(b) The angular velocity and linear velocity is connected by the relation,
- [tex]v = r \omega[/tex]
- Where, [tex]v[/tex] is the linear velocity, [tex]r[/tex] is the radius and [tex]\omega[/tex] is the angular velocity.
Substituting the known values, we get;
- [tex]v = (3\,m \times 2.22\,rad/s)=6.66 \, m/s[/tex]
(c) If the length of the sling is doubled, [tex]r = 6\,m[/tex].
This would change the angular displacement [tex]\theta[/tex] and hence the angular and linear velocities.
- [tex]A = \frac{\theta}{2\pi \,rad} \times \pi r^2[/tex]
- [tex]\implies \theta = \frac{A\times2\pi \,rad}{\pi r^2}= \frac{2A}{r^2} \,rad[/tex]
Substituting the known values, we get;
- [tex]\theta=\frac{2\times 50\, m^2}{36\,m^2}= 2.77 \, rad[/tex]
Therefore, the angular velocity will be,
- [tex]\omega =\frac{\theta}{t}=\frac{2.77\, rad}{5\,s} =0.554\,rad/s[/tex]
The linear velocity will be,
- [tex]v=r\omega = 6\,m \times 0.554\,rad/s = 3.324\,m/s[/tex]
- When the radius is doubled, the linear velocity is almost halved.
Learn more about rotational motion here:
https://brainly.com/question/14753185
