## Hovercraft B/C

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### Re: Hovercraft B/C

$P_1=\frac{mg}{t}=\frac{1*9.8}{1}=9.8W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Hope this is right   :oops:
Do the hovercraft fans power the centripetal force or does the tension in the wire?
Is it just 9.8 W?
Yep Of course, it would be more if I remembered to specify a coefficient of friction

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### Re: Hovercraft B/C

Yep Of course, it would be more if I remembered to specify a coefficient of friction
A uniform 0.4 m long stick rotates freely about a horizontal axis through one of its ends. It is released at an unknown angle to the vertical. When it hangs straight down, the speed of the tip of the stick is 3.5 m/s. How large is the unknown angle?
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But in ourselves, that we are underlings."

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### Re: Hovercraft B/C

A uniform 0.4 m long stick rotates freely about a horizontal axis through one of its ends. It is released at an unknown angle to the vertical. When it hangs straight down, the speed of the tip of the stick is 3.5 m/s. How large is the unknown angle?
Kind of confused here. Might be asking too much, but could you draw a picture?

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### Re: Hovercraft B/C

A uniform 0.4 m long stick rotates freely about a horizontal axis through one of its ends. It is released at an unknown angle to the vertical. When it hangs straight down, the speed of the tip of the stick is 3.5 m/s. How large is the unknown angle?
Kind of confused here. Might be asking too much, but could you draw a picture?
I think this picture shows what he's talking about:

The rod rotates in the plane of the page
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### Re: Hovercraft B/C

Ohh, I see.
$\frac12(\frac13m(0.4 m)^2)(3.5 \frac{m}{s} * \frac{1 rad}{0.4 m})^2 = mg(\frac12 * 0.4 m * cos \theta + \frac12 * 0.4 m)$
$\theta = 1.53 \textrm{radians}$
Not really sure since I still haven't quite wrapped my head around rotational stuff.

EDIT: Over-complicated things the first attempt

Justin72835
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### Re: Hovercraft B/C

Ohh, I see.
$\frac12(\frac13m(0.4 m)^2)(3.5 \frac{m}{s} * \frac{1 rad}{0.4 m})^2 = mg(\frac12 * 0.4 m * cos \theta + \frac12 * 0.4 m)$
$\theta = 1.53 \textrm{radians}$
Not really sure since I still haven't quite wrapped my head around rotational stuff.

EDIT: Over-complicated things the first attempt
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

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### Re: Hovercraft B/C

Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.

Justin72835
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Posts: 175
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### Re: Hovercraft B/C

Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.
Going from top to bottom, the tensions are as follows:

$Tension_1=(5+3+1)*9.8=88.2N$

$Tension_2=(3+1)*9.8=39.2N$

$Tension_3=(1)*9.8=9.8N$

Essentially, the tension in each rope is equal to the weight that it holds up.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

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### Re: Hovercraft B/C

Justin72835
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### Re: Hovercraft B/C

Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19