Rotational Question

  • Thread starter Paymemoney
  • Start date
  • #1
Paymemoney
175
0

Homework Statement


An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on the Moon is one-sixth that on the Earth. Radius of moon = 1.74 * 10^6 m
a) What must be the muzzle speed of the probe so that it travels completely around the Moon and returns to its original location? Found out to be 1.68km/s
b) How long does this trip around the Moon take?


Homework Equations


[tex]a=\omega^2r[/tex]

[tex]v=\omega r[/tex]

The Attempt at a Solution


I tired to solve with constant acceleration formulas.
a=1.62m/s^2
vi=0
vf=1.68km/s => 1678.928m/s

vf=vi+at
16878.928=0+1.62t
t=10419.09136s

2)A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

I have tried to do the following:

i know the static friction: 0.600

i need to know [tex]\omega[/tex]

which is [tex]\sqrt{9.8}{35}[/tex], [tex]\omega=0.529[/tex]

find velocity

[tex]v=r\omega[/tex]
v=0.529 * 35
v=18.52m/s

i know this is incorrect, someone tell me how i can approach this.
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
45,433
1,886

Homework Statement


An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on the Moon is one-sixth that on the Earth. Radius of moon = 1.74 * 10^6 m
a) What must be the muzzle speed of the probe so that it travels completely around the Moon and returns to its original location? Found out to be 1.68km/s
b) How long does this trip around the Moon take?


Homework Equations


[tex]a=\omega^2r[/tex]

[tex]v=\omega r[/tex]

The Attempt at a Solution


I tired to solve with constant acceleration formulas.
a=1.62m/s^2
vi=0
vf=1.68km/s => 1678.928m/s

vf=vi+at
16878.928=0+1.62t
t=10419.09136s
This is a circular motion/orbit problem. Use Newton's 2nd law and the formula for centripetal acceleration.
 
  • #3
Paymemoney
175
0
how can i find the time with the centripetal acceleration formula?
 
  • #4
Doc Al
Mentor
45,433
1,886
how can i find the time with the centripetal acceleration formula?
OK. Looks like you solved for the speed (but didn't show your work). The speed is constant. What distance does it travel in going around once?
 
  • #5
Paymemoney
175
0
do you mean [tex]\omega = 9.649 * 10^{4} rad/s[/tex]?
 
  • #6
Doc Al
Mentor
45,433
1,886
do you mean [tex]\omega = 9.649 * 10^{4} rad/s[/tex]?
You could certainly use that if you like. But that should be 10-4!

I was thinking of distance = speed * time. You have the speed. And you should be able to calculate the distance, then solve for the time.
 
  • #7
Paymemoney
175
0
How can i find the distance if i only have velocity, acceleration?
 
  • #8
Doc Al
Mentor
45,433
1,886
How can i find the distance if i only have velocity, acceleration?
What path does it take? Hint: Make use of the moon's radius.
 
  • #9
Paymemoney
175
0
would you find the diameter of the moon, which is 3480000?
 
  • #10
Doc Al
Mentor
45,433
1,886
Answer this: What is the shape of the package's orbit around the moon?
 
  • #11
Paymemoney
175
0
well the shape of a car....
 
  • #12
Doc Al
Mentor
45,433
1,886
well the shape of a car....
:bugeye:

It goes all the way around the moon in a big circle!
 
  • #13
Paymemoney
175
0
i misunderstood your question. -_-
 
  • #14
Doc Al
Mentor
45,433
1,886
Do you understand what I mean now? How can you find the length of that path? (Remember that this is a question about circular motion.)

(You could also solve for the time using ω, which you've already calculated.)
 
  • #15
Paymemoney
175
0
yes, i understand it took me awhile -_-.

So from the equation: [tex]\omega = \frac{\theta}{t}[/tex]

derived from the linear [tex]v=\frac{x}{t}[/tex]

where:

[tex]\theta = 2\pi[/tex] because it go around once.

[tex]\omega = 9.649 * 10^{-4}[/tex]


[tex]t=\frac{2\pi}{9.649*10^{-4}}[/tex]

t=6511.74765s

[tex]t=\frac{6511.74765}{3600}[/tex] turn into hours

t=1.81hrs however book's answer has a slight difference in 1.80hrs. Edit** i know why because of my rounding off.
 
  • #16
Paymemoney
175
0
would someone answer my second question?
 
Last edited:
  • #17
Doc Al
Mentor
45,433
1,886
yes, i understand it took me awhile -_-.

So from the equation: [tex]\omega = \frac{\theta}{t}[/tex]

derived from the linear [tex]v=\frac{x}{t}[/tex]
You used the first equation, but you could also have used the second one directly. When it goes around once, the package traces the circumference of a circle. So [itex]x = 2\pi r[/itex].
 
  • #18
Doc Al
Mentor
45,433
1,886
2)A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

I have tried to do the following:

i know the static friction: 0.600

i need to know [tex]\omega[/tex]

which is [tex]\sqrt{9.8}{35}[/tex], [tex]\omega=0.529[/tex]

find velocity

[tex]v=r\omega[/tex]
v=0.529 * 35
v=18.52m/s

i know this is incorrect, someone tell me how i can approach this.
How did you solve for ω?

Again, this is a circular motion problem that can be solved using Newton's 2nd law and what you know about centripetal acceleration.

What force acts on the eggs? What's the expression for its acceleration?

FYI: ac = ω²r = v²/r
 
  • #19
Paymemoney
175
0
How did you solve for ω?

Again, this is a circular motion problem that can be solved using Newton's 2nd law and what you know about centripetal acceleration.

What force acts on the eggs? What's the expression for its acceleration?

FYI: ac = ω²r = v²/r

the mass and the centripetal force.

Don't i need to use the static friction?
 
  • #20
Doc Al
Mentor
45,433
1,886
Don't i need to use the static friction?
Yes. Static friction is the only force available to provide the centripetal acceleration.
 
  • #21
Paymemoney
175
0
So would i start by finding the fsmax by using the equation

coefficient of static friction = [tex]\frac{fsmax}{n}[/tex]
 
  • #22
Doc Al
Mentor
45,433
1,886
So would i start by finding the fsmax by using the equation

coefficient of static friction = [tex]\frac{fsmax}{n}[/tex]
Right. The static friction will be at its maximum, so F = μN. (What's N?)

Now apply Newton's 2nd law.
 
  • #23
Paymemoney
175
0
the normal
 
  • #24
Paymemoney
175
0
how do i find the N???
 
  • #25
Doc Al
Mentor
45,433
1,886
how do i find the N???
To find an expression for the normal force, analyze the vertical force components. What's the net vertical force?
 
  • #26
Paymemoney
175
0
would it be F=N -mg
 
  • #27
Doc Al
Mentor
45,433
1,886
would it be F=N -mg
Right. And what must F equal in this case?
 
  • #28
Paymemoney
175
0
the static friction force
 
  • #29
Doc Al
Mentor
45,433
1,886
the static friction force
No. We're talking about vertical forces here. What's the net vertical force? (Is there any vertical acceleration?)

The purpose is just so you can find the normal force.
 
  • #30
Paymemoney
175
0
wouldn't that be mg???
 
  • #31
Doc Al
Mentor
45,433
1,886
  • #32
Paymemoney
175
0
but how do i know the mass??
 
  • #33
ideasrule
Homework Helper
2,284
0
You don't need it. Use call the mass "m" and you'll see it cancels out.
 
  • #34
Paymemoney
175
0
so if i use the F= N - mg

mg = N - mg

9.8 = N -9.8

N= 19.6N
 
  • #35
Doc Al
Mentor
45,433
1,886
but how do i know the mass??
You don't and it doesn't matter. Just use N = mg.

so if i use the F= N - mg

mg = N - mg

9.8 = N -9.8

N= 19.6N
No. The vertical forces add to zero, so 0 = N - mg, thus N = mg.

Now go back to analyzing the friction force. Now that you have an expression for N, how would you express the maximum static friction? Plug that into Newton's 2nd law.
 

Suggested for: Rotational Question

Replies
9
Views
358
  • Last Post
Replies
10
Views
488
Replies
9
Views
671
Replies
1
Views
253
  • Last Post
Replies
9
Views
464
Replies
4
Views
338
Top