185k views
4 votes
5/53 The circular disk of radius 8 in. is released very near the horizontal surface with a velocity of its center vO = 27 in./sec to the right and a clockwise angular velocity ω = 2 rad/sec. Determine the velocities of points A and P of the disk. Describe the motion upon contact with the ground.,

1 Answer

5 votes

Answer:

vA = 27 i in/s + 16 j in/s

vA = 31.38 in/s

vP = 11 i in/s

vP = 11 in/s

Step-by-step explanation:

We write the known values in the vector form using the right hand rule.

Given

ω = - 2 k rad/s

v₀ = 27 i in/s

For point A

vA = v₀ + ωxR_OA

vA = 27 i in/s + (- 2 k rad/s)x(- 8 i in)

vA = 27 i in/s + 16 j in/s

then we get the module as follows

vA = √((27 in/s)² + (16 in/s)²)

⇒ vA = 31.38 in/s

For point P

vP = v₀ + ωxR_OP

vP = 27 i in/s + (- 2 k rad/s)x(- 8 j in)

vP = 27 i in/s - 16 i in/s

vP = 11 i in/s

then, the module is

vP = 11 in/s

The pic shown can help us to understand the question.

5/53 The circular disk of radius 8 in. is released very near the horizontal surface-example-1
User Leydi
by
5.5k points