A) The speed of the tip of the blade is 76.2 m/s
B) The centripetal acceleration of the tip of the blade is [tex]103.7 m/s^2[/tex]
D) The force required to keep the droplet moving in circular motion is 0.39 N
E) The ratio of the force to the weight of the droplet is 4.0
Explanation:
A)
We know that the blade of the turbine is rotating at an angular speed of
[tex]\omega = 13 rpm[/tex]
First, we have to convert this angular speed into radians per second. Keeping in mind that
[tex]1 rev = 2 \pi[/tex]
[tex]1 min = 60 s[/tex]
We get
[tex]\omega = 13 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=1.36 rad/s[/tex]
The linear speed of a point on the blade is given by:
[tex]v=\omega r[/tex]
where
[tex]\omega=1.36 rad/s[/tex] is the angular speed
r is the distance of the point from the axis of rotation
For a point at the tip of the blade,
r = 56 m
Therefore, its speed is
[tex]v=(1.36)(56)=76.2 m/s[/tex]
B)
The centripetal acceleration of a point in uniform circular motion is given by
[tex]a=\frac{v^2}{r}[/tex]
where
v is the linear speed
r is the distance of the point from the axis of rotation
In this problem, for the tip of the blade we have:
v = 76. 2 m/s is the speed
r = 56 m is the distance from the axis of rotation
Substituting, we find the centripetal acceleration:
[tex]a=\frac{(76.2)^2}{56}=103.7 m/s^2[/tex]
D)
The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:
[tex]F=m\frac{v^2}{r}[/tex]
where
m is the mass of the droplet
v is the linear speed
r is the distance from the centre of rotation
The data in this problem are
m = 10 mg = 0.010 kg is the mass of the droplet
v = 2.5 m/s is the linear speed
r = 16 cm = 0.16 m is the radius of the circular path
Substituting,
[tex]F=(0.010)\frac{2.5^2}{0.16}=0.39 N[/tex]
E)
The weight of the droplet is given by
[tex]F_g = mg[/tex]
where
m = 10 mg = 0.010 kg is the mass of the droplet
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting,
[tex]F_g = (0.010)(9.8)=0.098 N[/tex]
The force that keeps the droplet in circular motion instead is
F = 0.39 N
Therefore, the ratio between the two forces is
[tex]\frac{F}{F_g}=\frac{0.39}{0.098}=4.0[/tex]
Learn more about circular motion:
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