Complete Question:
A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.
Answer:
a) [tex]v_{A} = 10 m/s[/tex]
b)
[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]
Explanation:
a) Velocity of point A
The velocity of point A is a combination of the translational and the rotational velocity of the racket.
[tex]v_{A} = v_{G} + v_{r}[/tex]
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = wr[/tex]
r = 25 cm = 0.25 m
w = 20 rad/s
[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]
[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]
b) Velocity of point B
At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]
[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]
In the y - direction, r = 30/2 = 15 cm = 0.15 m
r = 0.15 m
[tex]w_{x} = 1 rad/s[/tex]
[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]
In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]
[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]
[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s
