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Angular momentum question

  • #1

Homework Statement


A small object with mass 3.20 kg moves counterclockwise with constant speed 6.30 m/s in a circle of radius 4.70 m centered at the origin. It starts at the point with position vector (4.70 i+ 0j ) m. Then it undergoes an angular displacement of 8.00 rad.
What it its position vector?
In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
What is its velocity?
What is its acceleration?
What total force is exerted on the object?


Homework Equations


i believe its the kinematics equations just replaced with angular speed and acceleration variables



The Attempt at a Solution


honestly..i think i should know how to do this but i cant seem to figure it out
any lil start would help..thanks
 

Answers and Replies

  • #2
Doc Al
Mentor
44,904
1,169
How many radians in a circle?
 
  • #3
umm isnt like 2pi radians
 
  • #4
Doc Al
Mentor
44,904
1,169
umm isnt like 2pi radians
Yes. Use that fact to figure out where the object ends up on the circle after it moves 8.00 radians.
 
  • #5
ok so it goes around 4 times i guess?
i still dont understand
 
  • #6
1,013
70
8 much less than 4*2Pi.
 
  • #7
ok so i did 8rad/2pi = 1.27
what do i do with this information?
 
  • #8
1,013
70
So the object has undergone 1 full revolution and .27 of a full revolution, which means that the rotation of the object from its original location is .27*2Pi radians. Can you see why and how to get the resulting vector? It is very important to draw a picture since this is the first time you are doing this.
 
  • #9
hmm ok so im pretty sure this means the particle is in the second quadrant right?
so .27*2Pi = 1.70...im not quite sure how to get the resulting vector tho...is it just 4.7*1.70?
 
  • #10
1,013
70
hmm ok so im pretty sure this means the particle is in the second quadrant right?
so .27*2Pi = 1.70...im not quite sure how to get the resulting vector tho...is it just 4.7*1.70?
4.7 is the length of the vector along the x-axis, while 1.7 is the angle the vector we are looking for makes with the x-axis. Why are you multiplying a length with an angle?
Draw a picture. We want to rotate the vector 4.7i + 0j, which luckily lies along the x-axis, 1.7 radians. The length of the vector is 4.7, so by the definition of sine and cosine, this gives us 4.7*sin(1.7) for the y-component and 4.7*cos(1.7) for the x-component. Remember that the 1.7 is in radians, not degrees. You will have to translate that if you are using a calculator that only works in degrees.
If you have drawn a diagram, you might also notice that this is the same as 4.7*sin(Pi - 1.7) and -4.7*cos(Pi - 1.7).
 
  • #11
wow okay that makes so much more sense
so for x component = -.61
y component = 4.67

how do i do part b
i think part c for xcomponent its -6.30 m/s but i dont know the y component
 
  • #12
1,013
70
how do i do part b
You already know the angle it makes from part a. From the signs of the components, or from your knowledge of angles, you should know the quadrant.
i think part c for xcomponent its -6.30 m/s but i dont know the y component
Draw a diagram. The tangential velocity vector starts at the tip of the position vector for the particle, is perpendicular to the position vector, and you have its length of 6.3. Getting the components is geometry/trigonometry similar to what we did in part a.
 
  • #13
ok cool i think i got those parts anyways..but thanks for explaining them

how about e and f..any hints
 
  • #14
1,013
70
You need to use the definition of centripetal force.
 
  • #15
o ok..well that means that a = v^2/r
so 6.3^2/4.7 = 8.44m/s^2 right..how do i put that in x and y components

and for f its just mv^2/r right?
 

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