Uniform Circular Motion: Finding Velocity in Unit-Vector Notation

[HeRo]

I thought I had this problem on lock, but apparently I was wrong in my methods. My original answer was 7.97i + 3.92j but that is incorrect. I've tried other methods but now I'm just lost. I need some assistance or I'm a dead man.

"A particle is in uniform circular motion about the origin of an xy coordinate system, moving counterclockwise with a period of 8.00 s. At one instant, its position vector (from the origin) is r = (6.00 m) i - (5.00 m) j. At that instant, what is its velocity in unit-vector notation?"

 

[Tide]

There are several ways of approaching this one.

Since you know the position at one instant and the particle is orbiting the origin you can determine the radius of the orbit and, therefore, the circumference of the orbit. Divide the circumference by the period to find the speed.

Now just find a unit vector in the plane and perpendicular to the position vector, multiply by the speed and you have the result you need. Just be sure to pick the unit vector so that the motion is counterclockwise.

A second approach is to write the position vector as
\vec r = r (\cos (\omega t + \phi) \hat i + \sin (\omega t + \phi) \hat j
and use the data provided to determine \phi and r.

 

[wais]

It's understandable to feel frustrated when working on a problem and not getting the correct answer. It's important to remember that mistakes are a normal part of the learning process and seeking assistance when needed is a great way to improve. Let's take a look at the problem and see if we can figure out where the mistake might be.

First, let's review the steps for finding velocity in unit-vector notation for uniform circular motion. We know that the velocity vector is tangent to the circular path and its magnitude is equal to the speed of the particle. We also know that the direction of the velocity vector is perpendicular to the position vector at any given point on the circular path. This means that the velocity vector will always be pointing towards the center of the circle.

Now, let's look at the given information. We know that the particle has a period of 8.00 s, which means it completes one full revolution in that time. This also means that its speed is constant throughout the motion. We are given the position vector at one instant, which is r = (6.00 m) i - (5.00 m) j. This means that the particle is 6.00 m away from the origin in the x-direction and 5.00 m away in the y-direction.

To find the velocity vector, we can use the formula v = (2πr)/T, where r is the radius and T is the period. In this case, the radius is the magnitude of the position vector, which is √(6.00^2 + 5.00^2) = 7.81 m. Plugging in the values, we get v = (2π)(7.81 m)/8.00 s = 4.87 m/s. This is the magnitude of the velocity vector.

Next, we need to find the direction of the velocity vector. As mentioned earlier, it will always be pointing towards the center of the circle, which in this case is the origin. This means that the velocity vector will have components in both the x and y directions. We can use the ratio of the sides of a right triangle to find the components. In this case, the x-component will be 6.00 m and the y-component will be 5.00 m. Therefore, the velocity vector in unit-vector notation will be v = (6.00 m/s) i + (5.00 m/s) j. This