I can't do space curvilinear motion

In summary, the book says that the acceleration in the y direction is 23m/s^2, while the equation in the book says that the acceleration in the y direction is 34.4m/s^2.
  • #1
EddieHimself
9
0

Homework Statement



[PLAIN]http://a4.sphotos.ak.fbcdn.net/hphotos-ak-ash4/s720x720/300201_1919471514513_1473790586_31632884_1692676_n.jpg

Homework Equations



aR = [itex]\ddot{R}[/itex] - R[itex]\dot{\phi}[/itex]2 - R[itex]\dot{\theta}[/itex]2cos2([itex]\phi[/itex])

a[itex]\theta[/itex] = [itex]\frac{cos(\phi)}{R}[/itex][itex]\frac{d}{dt}[/itex](R2[itex]\dot{\theta}[/itex]) - 2R[itex]\dot{\theta}\dot{\phi}[/itex]sin[itex]\phi[/itex]

a[itex]\phi[/itex] = [itex]\frac{1}{r}[/itex][itex]\frac{d}{dt}[/itex](R2[itex]\dot{\phi}[/itex]) + R[itex]\dot{\theta}[/itex]2sin[itex]\phi[/itex]cos[itex]\phi[/itex]

The Attempt at a Solution

 
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  • #2
You can calculate y,tt (the acceleration in the y direction) = v^2/r = 166*166/1200 = 23m/s^2 the acceleration in the y direction. With this you can find a_r and a_phi by some trig. You know at the instant of calculation that phi,t = 0, R,t = 0, and using some trig you can calculate theta,t.

Does this help?
 
  • #3
i've already done that bit. I worked out the normal acceleration for the loop, [itex]\dot{\theta}[/itex] as 0.1437 rad/s, ar = 8.601 m/s2 and a[itex]\phi[/itex] as 21.49 m/s2 but the problem is when i try to input all this into working out the values of [itex]\ddot{R}[/itex] and [itex]\ddot{\phi}[/itex] that i seem to come out with something completely different.
 
  • #4
The units you have for a_phi and the units for a_phi in the answer are different. I think you are off by a factor of R?
 
  • #5
Spinnor said:
The units you have for a_phi and the units for a_phi in the answer are different. I think you are off by a factor of R?

It's asking me to work out [itex]\ddot{\phi}[/itex] (rad/s2), which is a different quantity to a[itex]\phi[/itex], (m/s2). If i just divide 21.49/1077 that equals 0.0199 which is not the right answer. My problem is that i don't know whether the value i have for a[itex]\phi[/itex] is wrong, or whether I've done something wrong in the equation or what?
 
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  • #6
Maybe I'm doing something wrong but I suspect the answer in the book. I got an acceleration in the y direction of about 23m/s^2 and R,tt should be a fraction of that number and not the 34.4m/s^2 in the book. I'm stumped, if you get an answer please let us know, if I'm doing something wring I would like to know.
 

What is space curvilinear motion?

Space curvilinear motion refers to the movement of an object in three-dimensional space along a curved path. This type of motion involves both linear and angular motion.

What causes an object to experience space curvilinear motion?

Space curvilinear motion can be caused by various factors such as gravitational force, centripetal force, or external forces acting on the object. The motion may also be a result of the object's own inertia.

How is space curvilinear motion different from linear motion?

Unlike linear motion, space curvilinear motion involves movement along a curved path rather than a straight line. This type of motion also involves a combination of linear and angular motion, making it more complex.

What are some real-life examples of space curvilinear motion?

Some examples of space curvilinear motion include the orbit of planets around the sun, the motion of a rollercoaster on its track, and the flight path of a bird.

How is space curvilinear motion relevant to science and engineering?

Understanding space curvilinear motion is crucial in fields such as physics and engineering. It helps scientists and engineers analyze and predict the movement of objects in three-dimensional space, which is essential in designing various structures and technologies.

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