A particle P travels with constant speed on a circle of radius r = 2.50 m and co

Question

A particle P travels with constant speed on a circle of radius r = 2.50 m and completes one revolution in 20.0 s. The particle passes through O at time t = 0. State the following vectors in magnitude-angle notation (angle relative to the positive direction of x).

A particle P travels with constant speed on a circle of radius r = 2.50 m and completes one revolution in 20.0 s. The particle passes through O at time t = 0. State the following vectors in magnitude-angle notation (angle relative to the positive direction of x). With respect to O, find the particle's position vector at the time t = 5.00 s. With respect to O, find the particle's position vector at the time t = 7.50 s. With respect to O, find the particle's position vector at the time t = 10.00 s. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's average velocity. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's velocity at the beginning. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's velocity at the end of the 5.00 s interval. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the acceleration at the beginning. For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the acceleration at the end of the interval.

Explanation / Answer

First note that we're dealing with uniform circular motion since the particle is moving along a circular path at constant speed. Therefore all acceleration will be directed towards the center of the circular path because if this wasn't the case the speed would not be constant and you would have a tangental component of acceleration.

At t1=2s we have (R is radius of circle):
a = v^2 / R = (5.00)^2 / R = 25/R m/s^2 in positive x direction
Note that the velocity cannot have an x component because that would make the motion nonuniform, the question should say "with velocity (5.00 m/s) in positive y-direction".

At t2=6s we have:
a = v^2 / R = 25/R m/s^2 in positive y direction
Same as before, but here velocity is (-5.00 m/s) in positive x-direction. Empasis on negative 5.

Since the acceleration always points towards the center of the circle we can deduce from the information given at t1:
center = (x, 3.00)
where x>2.00m

Now the particle will have gone through 3/4ths of a rotation from t1 to t2. Think about why this is so. First lets find the linear distance traveled in that time:
distance = 5.00m/s * (6-2 s) = 20m
We know that the particle have traveled 3/4ths of the way around the circular path so the distance can also be given in terms of the radius R:
distance = 2pi*(3/4)*R
Note that the circumference of a circle is 2pi*R so 3/4ths of the circumference is the equation given. Equating the two yields:
20 = 2pi*(3/4)*R
40/(3pi) = R

The x-coordinate is therefore 2.00+40/(3pi), making the center of the path
(2.00+40/(3pi)m , 3.00m)

Edit: Also note that I never used the equations for the acceleration. They aren't necessary so if you're wondering why they're there just ignore them. Just a habit of mine (writing down too much information doesn't hurt).

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