A particle travels counterclockwise around the origin at constant speed in a circular path that has a diameter of 2.50 m, going around twice per second. Point Q is on the path halfway between points P and R, which lie on the x axis and the y axis, respectively. State the following velocites as vectors in polar notation. (a) What is the particle's average velocity over the interval PR? (b) What is its average velocity over the interval PQ? (c) What is its instantaneous velocity at point P?
P coordinates: (1.25, 0)
R coordinates: (0, 1.25)
Q is directly between the two
frequency f = 2 rev/s
angular velocity = 2•π•f = 4•π rad/s
(a) Velocity vector at P = (1.25x4x3.14)j = 15.7 j,
where j is unit vector along positive y axis.
Also velocity vector of the particle at R = -15.7i,
where i is unit vector along positive x direction.
As the direction changes uniformly,
the average velocity over the interval PR is
(15.7/2)(j-i)= 7.85(j-i).
Magnitude = sqrt[7.85^2+7.85^2] =
=1.414x7.85 = 11.10 m/s
and the direction is 45 degrees west of north.
(b) Average velocity, V1 over interval PQ is
= 0.5{15.7 j + [15.7/sqrt(2)](j-i)} or
V1 = 0.5[26.80 j - 11.10i] = 13.40 j - 5.55 i.
Magnitude of V1 = sqrt[13.40^2 + 5.55^2] = 14.50 m/s.
due theta west of north given by theta = tan^-1 [5.55/13.40] = 22.5 degrees west of north.
(c) v= 2•π•f•R = 15.7 m/s due north
To find the velocities as vectors in polar notation, we need to first find the displacements and time intervals for the given intervals. Let's start by calculating the necessary values.
Given information:
- Particle travels counterclockwise around the origin.
- Circular path diameter = 2.50 m.
- The particle completes two revolutions per second.
(a) Average velocity over the interval PR:
To find the displacement for interval PR, we need to calculate the change in x-coordinate and change in y-coordinate.
Change in x-coordinate (Δx):
Δx = xf - xi = 0 - 1.25 = -1.25 m
Change in y-coordinate (Δy):
Δy = yf - yi = 1.25 - 0 = 1.25 m
Now, let's find the time interval for interval PR:
The particle completes two revolutions per second, which means it takes 0.5 seconds for one revolution.
Time interval (Δt):
Since the particle goes from P to R, it covers a quarter of a revolution.
Δt = (1/4) * (1/2 s) = 1/8 s
Now, we can calculate the average velocity over the interval PR using the displacement and time interval.
Average velocity over interval PR (v_avg PR):
v_avg PR = Δd/Δt,
where Δd is the displacement and Δt is the time interval.
v_avg PR = (Δx/Δt, Δy/Δt)
Substituting the values,
v_avg PR = (-1.25 / (1/8), 1.25 / (1/8))
= (-10 m/s, 10 m/s)
Therefore, the particle's average velocity over the interval PR in polar notation is (-10 m/s, 10 m/s).
(b) Average velocity over the interval PQ:
Since point Q is directly between points P and R, interval PQ is half of interval PR.
Displacement for interval PQ is half of Δx and Δy from interval PR:
Change in x-coordinate for interval PQ (Δx_PQ) = Δx / 2
= -1.25 / 2
= -0.625 m
Change in y-coordinate for interval PQ (Δy_PQ) = Δy / 2
= 1.25 / 2
= 0.625 m
Time interval for interval PQ is half of Δt from interval PR:
Δt_PQ = Δt / 2
= (1/8) / 2
= 1/16 s
Using these values, we can calculate the average velocity over the interval PQ:
Average velocity over interval PQ (v_avg PQ) = (Δx_PQ / Δt_PQ, Δy_PQ / Δt_PQ)
Substituting the values,
v_avg PQ = (-0.625 / (1/16), 0.625 / (1/16))
= (-10 m/s, 10 m/s)
Therefore, the particle's average velocity over the interval PQ in polar notation is (-10 m/s, 10 m/s), which is the same as the average velocity for interval PR.
(c) Instantaneous velocity at point P:
The instantaneous velocity at a given point on a circular path can be found by considering the direction and magnitude of the tangential velocity at that point.
At point P, the particle is moving counterclockwise. The magnitude of the velocity is given by the speed at which the particle travels around the circular path.
Given that the particle completes two revolutions per second, and the radius of the circular path (half the diameter) is 2.50/2 = 1.25 m, we can find the speed:
Speed = 2πr/T,
where r is the radius and T is the time for one revolution.
Speed = 2π(1.25)/(1/2)
= 10π m/s
The instantaneous velocity at point P is in the counterclockwise direction and has a magnitude of 10π m/s.
Therefore, the particle's instantaneous velocity at point P in polar notation is (10π m/s, 90°), indicating a magnitude of 10π m/s and an angle of 90 degrees counterclockwise from the positive x-axis.