a) The angular speed is 136.1 rad/s
b) The tangential speed is 3.97 m/s
c) The radial acceleration is [tex]1493 m/s^2[/tex]
d) The total distance covered by a point on the rim in 1.94 s is 21.3 m
Explanation:
a)
We want to determine the angular speed of the disk, which is given by the ratio between the angular displacement and the time taken:
[tex]\omega = \frac{\theta}{t}[/tex]
where [tex]\theta[/tex] is the angular displacement and t the time taken.
In this problem, we are already given the angular speed, which is
[tex]\omega=1300 rev/min[/tex]
We need to convert it into rad/sec. Keeping in mind that
[tex]1 rev = 2\pi rad\\1 min = 60 s[/tex]
we have
[tex]\omega=1300 \frac{rev}{min}\cdot \frac{2\pi rad/rev}{60 s/min}=136.1 rad/s[/tex]
b)
The tangential speed of a point on the disk is given by
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the point from the axis of rotation
In this problem, we have
[tex]\omega=136.1 rad/s[/tex]
r = 2.92 cm = 0.0292 m
Substituting,
[tex]v=(136.1)(0.0292)=3.97 m/s[/tex]
c)
The radial acceleration (also called centripetal acceleration) of a point on the disk is given by
[tex]a_c = \omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the point from the axis of rotation
Here we have:
[tex]\omega=136.1 rad/s[/tex]
r = 8.06 cm = 0.0806 m is the distance of a point on the rim from the axis
Substituting,
[tex]a_c = (136.1)^2(0.0806)=1493 m/s^2[/tex]
d)
The total distance covered by a point on the rim in a certain time t is given by
[tex]d=vt[/tex]
where
v is the tangential speed of a point on the rim
t is the time
Since the tangential speed can be rewritten as [tex]v=\omega r[/tex], the previous equation becomes
[tex]d=\omega r t[/tex]
where we have:
[tex]\omega=136.1 rad/s[/tex]
r = 8.06 cm = 0.0806 m
t = 1.94 s
Substituting,
[tex]d=(136.1)(0.0806)(1.94)=21.3 m[/tex]
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