Uniform circular motion of a particle problem

In summary, the particle is moving along a circular path at constant speed in a horizontal xy coordinate system. At time t1 = 3.30 s, it is at point (5.10 m, 6.80 m) with a velocity of (3.90 m/s)## \hat j ## and acceleration in the positive x direction. At time t2 = 9.90 s, it has a velocity of (–3.90 m/s)## /hat i ## and acceleration in the positive y direction. To determine the coordinates of the center of the circular path, the circumference and radius must be calculated using the tangential speed and time traveled. The particle travels 3/4 around the circle in a
  • #1
J-dizzal
394
6

Homework Statement


A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.30 s, it is at point (5.10 m, 6.80 m) with velocity (3.90 m/s)## \hat j ## and acceleration in the positive x direction. At time t2 = 9.90 s, it has velocity (–3.90 m/s)## /hat i ##and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.

Homework Equations


a=v2/r, v=2πr/T, [/B]

The Attempt at a Solution


hMthL4b3OjGjbIWmGkYHA%2F12%2F424149441%2Fjpeg%2F32x32%2F1%2F1435046400%2F0%2F2%2F20150622_205717.jpg
 
Physics news on Phys.org
  • #2
If all its speed is in j-hat while at (5.1, 6.8), then it is crossing the horizontal axis through the circle at that moment. Hence, the y-component of the center is the y-component of (5.1,6.8)

It’s tangential speed is 3.9 m/s. It takes 9.9-3.3 seconds to travel ¼ around its orbit. (I’m assuming it didn’t go 5/4, or 9/4 around, etc.). So now you have a speed and a time, so you can solve for distance (circumference). Then it’s easy to get radius. If you know the radius and you know its position when all its velocity in is the j hat direction, then your radius is the offset from you x-component of that position.
 
  • #3
tony873004 said:
If all its speed is in j-hat while at (5.1, 6.8), then it is crossing the horizontal axis through the circle at that moment. Hence, the y-component of the center is the y-component of (5.1,6.8)

It’s tangential speed is 3.9 m/s. It takes 9.9-3.3 seconds to travel ¼ around its orbit. (I’m assuming it didn’t go 5/4, or 9/4 around, etc.). So now you have a speed and a time, so you can solve for distance (circumference). Then it’s easy to get radius. If you know the radius and you know its position when all its velocity in is the j hat direction, then your radius is the offset from you x-component of that position.
would'nt it travel 3/4 around the circle 3pi/2 if its velocity is positive jhat at the left side of the circle its moving clockwise?
 
  • #4
Is it given that its moving clockwise?
 
  • #5
tony873004 said:
Is it given that its moving clockwise?
no but both vectors are pointing in a direction that would indicate clockwise motion
 
  • #6
You're right, it moves 3/4 around.
 

FAQ: Uniform circular motion of a particle problem

1. What is uniform circular motion?

Uniform circular motion refers to the motion of an object or particle that moves in a circular path at a constant speed. This means that the object is moving at the same speed throughout its circular path, but its direction is constantly changing.

2. How is the velocity of a particle calculated in uniform circular motion?

The velocity of a particle in uniform circular motion is calculated by dividing the circumference of the circle by the time it takes for the particle to complete one full revolution. This is known as the tangential velocity, and it is always perpendicular to the radius of the circle at any given point.

3. What is the centripetal force in uniform circular motion?

The centripetal force is the force that acts towards the center of the circle, keeping the particle in its circular path. It is equal to the mass of the particle multiplied by its tangential velocity squared, divided by the radius of the circle.

4. How is the period of a particle in uniform circular motion related to its frequency?

The period of a particle in uniform circular motion is the time it takes for the particle to complete one full revolution. The frequency, on the other hand, is the number of revolutions the particle completes in one second. The two are inversely related, meaning that as the period increases, the frequency decreases and vice versa.

5. Can the velocity of a particle in uniform circular motion change?

No, the velocity of a particle in uniform circular motion remains constant throughout its circular path. However, its direction changes, which means that its velocity vector is constantly changing. This is due to the centripetal force acting on the particle, which causes it to continuously change direction.

Similar threads

Acceleration in uniform circular motion is uniform or non-uniform?
2
Replies
55
Views
2K
Particle Uniform circular motion problem
Replies
4
Views
1K
Would particle P be under non-unform circular motion?
Replies
29
Views
1K
Relation between the velocity vector and the acceleration vector of an object
Replies
1
Views
201
Find the acceleration in circular motion
Replies
6
Views
1K
Uniform Circular Motion - distance
Replies
2
Views
1K
Circular motion to projectile motion
Replies
18
Views
2K
Circular Motion Questions (energies, forces, angular velocities, etc.)
Replies
2
Views
977
Uniform Circular Motion Inside a Sphere of Charge
Replies
7
Views
3K
Is Vertical Circular Motion Ever Uniform?
Replies
10
Views
2K

Hot Threads

Recent Insights

Back
Top