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?
Point P is at coordinates (1.25,0) and point R is at coordinates (0,1.25). Point Q is directly in the middle of P and R. Can you explain the answers as well so I understand?!
Part C starts with (15.7, ? degrees)
That's all I know.
Linear velocity is v = ω•R=2•π•n•R =2•π•2•1.25 =15.7 m/s.
The quarter of the circle is covered for the time
t = 2•π•R/4•2•π•n•R =1/8 s =0.125 s.
Displacement for this time is the distance betwee point P and Q
D =sqrt(R² +R²) 1.25•1.41 = 1.76m.
v(ave) =displacement/ time =
1.76/0.125 =14.1 m/s.
(b)
Time for covering the 1/8 part of circle is
t =1/16 =0.0625 s.
Displacement (using cosine law) is
D =sqrt(R² +R² -2 R²cos45º)=0.96 m
v (ave) = 0.96/0.0625 =15.3.
(c)
Instanteneous linear velocity is
v = ω•R=2•π•n•R =2•π•2•1.25 =15.7 m/s.
To find the average velocity over a given interval, we need to determine the displacement vector, which is the vector pointing from the initial position to the final position of the particle.
(a) To find the average velocity over the interval PR, we need to calculate the displacement vector PR.
The displacement vector PR can be found by subtracting the initial position vector P from the final position vector R.
PR = R - P = (0, 1.25) - (1.25, 0) = (-1.25, 1.25)
The average velocity is the displacement vector divided by the time interval. Since the particle goes around twice per second, the time interval is 1/2 second.
Average velocity over PR = PR / (1/2) = (-1.25, 1.25) / (1/2) = (-2.5, 2.5) m/s
So, the average velocity over interval PR is (-2.5, 2.5) m/s in polar notation.
(b) To find the average velocity over the interval PQ, we need to calculate the displacement vector PQ.
The displacement vector PQ can be found by subtracting the initial position vector P from the final position vector Q.
PQ = Q - P = (1.875, -1.875) - (1.25, 0) = (0.625, -1.875)
The average velocity over PQ = PQ / (1/2) = (0.625, -1.875) / (1/2) = (1.25, -3.75) m/s
So, the average velocity over interval PQ is (1.25, -3.75) m/s in polar notation.
(c) To find the instantaneous velocity at point P, we need to find the velocity vector at that point. The velocity vector is a tangent to the circular path of the particle at point P.
Since the particle travels counterclockwise around the origin at a constant speed, the velocity vector at any point on the circular path is tangent to the direction of motion and points in the direction of increasing polar angle (θ).
The angle (θ) can be found by considering the position vector at point P (1.25, 0) in polar notation. The angle θ can be calculated as:
θ = arctan(y/x) = arctan(0/1.25) = 0 degrees
The magnitude of the velocity vector can be determined by considering the speed of the particle. Since it goes around twice per second, the magnitude is 2 times the radius (diameter/2).
Magnitude of velocity vector = 2 * (2.5/2) = 2.5 m/s
Therefore, the instantaneous velocity at point P can be written in polar notation as (2.5, 0 degrees).
Note: In polar notation, the velocity vector is represented as (magnitude, direction in degrees).