Motion in 2d/Centripetal Acceleration

So in summary, we used kinematic equations and trigonometry to find the x and y components of velocity, coordinates, and speed of the particle in the first scenario. In the second scenario, we used trigonometry and the formula for centripetal acceleration to find the magnitude and direction of the ball's velocity.
  • #1
HurricaneH
10
0
...sorry for doing this, but having trouble with some problems

1) A particle starts from rest at t=0 at the origin and moves in the xy plane with a constant acceleration of a= (2i + 4j) m/s2. After a time t has elaped, determine (a) the x and y components of velocity. (b) the coordinates of the particle, and (c) the speed of the particle.

I get a) Vx= (2t)m/s and Vy=(4t)m/s, but not the rest...


and...

2) A ball swings in a vertical circle at the end of a rope 1.5 M long. When it is 36.9* past the lowest point on its way up, the ball's total acceleration is (-22.5i + 20.2j) m/s2. For that instant, a) determine the magnitude of its centripetal acceleration, and b) determine the magnitude and direction of its velocity.

I know that v^2/R= Ac, but i don't know how to get the velocity:(

Any help will be appreciated
 
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  • #2
1) A particle starts from rest at t=0 at the origin and moves in the xy plane with a constant acceleration of a= (2i + 4j) m/s2. After a time t has elaped, determine (a) the x and y components of velocity. (b) the coordinates of the particle, and (c) the speed of the particle.

Use uniform acceleration kinematic equations, or use calculus. Integrate Acceleration, then integrate Velocity.

[tex] \vec{a} = \frac{d \vec{v}}{dt} [/tex]

[tex] \vec{v} = \frac{d \vec{r}}{dt} [/tex]


2) A ball swings in a vertical circle at the end of a rope 1.5 M long. When it is 36.9s past the lowest point on its way up, the ball's total acceleration is (-22.5i + 20.2j) m/s2. For that instant, a) determine the magnitude of its centripetal acceleration, and b) determine the magnitude and direction of its velocity

Remember

[tex] A_{c} = \frac{v^2}{r} [/tex]

Refers to Magnitude.
 
  • #3
but how do i find out the velocity?
 
  • #4
Well I got it. For 1 b) x=t^2m y=2t^2, for c) a^2 + b^2= c^2, c= velocity.

For 2. The angles of the directionals were 36.9 and 53.1, then cos36.9=20.2/x, solve for x, cos53.1= 22.5/x2, solve for x2, add both xs, and you get 62.8 m/s for the magnitude of its centripetal acceleration. For c, v^2/r= aC, we have the aC and r, so v^2/1.5=62.8, simple algebra, and v=9.7m/s.
 

Related to Motion in 2d/Centripetal Acceleration

1. What is the difference between motion in 2D and motion in 3D?

Motion in 2D refers to the movement of an object along two axes, typically represented as horizontal and vertical. Motion in 3D involves movement along three axes, such as forward/backward, left/right, and up/down. In 2D motion, the object's position can be described with two coordinates, while in 3D motion, three coordinates are needed.

2. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is caused by the centripetal force, which keeps the object moving in a circular path.

3. How is centripetal acceleration calculated?

The formula for centripetal acceleration is a = v^2/r, where "a" is the centripetal acceleration, "v" is the velocity of the object, and "r" is the radius of the circular path.

4. What is the relationship between centripetal acceleration and centripetal force?

Centripetal force is the force that is responsible for keeping an object moving in a circular path. It is directly proportional to the centripetal acceleration, meaning that as the centripetal force increases, so does the centripetal acceleration. This relationship can be expressed as F = ma = mv^2/r.

5. Can an object have centripetal acceleration without a centripetal force?

No, an object cannot have centripetal acceleration without a centripetal force. According to Newton's first law of motion, an object will continue to move in a straight line at a constant speed unless acted upon by a force. In the case of circular motion, the centripetal force is necessary to change the object's direction and keep it moving in a circular path.

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