A circular plate with a slot in its surface is rotating with angular velocity ω and angular acceleration α with respect to a fixed coordinate system, OXYZ. A ball B moves along the slot in the direction shown with velocity, v, and acceleration, a, both measured relative to the plate. A body coordinate system Axyz is attached to the rotating plate , i^ runs parallel to the slot and at this instant the ball is located on the j^ unit vector at a length of L from the center of the disk A. The length of the slot is 2L. The ball is slightly smaller than the sot and therefore only contacts one side of the slot at at time.
-A is the point at the center of the disk. The radius of AB is L and located on the unit vector j^.
-j^ splits down the middle of OXYZ with 45 degrees on each side.
QUESTIONS:
1)For the instant of time described , give the ball's velocity vB with respect to the fixed ground frame Oxyz. Express your answer using the body coordinates i^ and j^.
Please enter the quantity terms of a,v,α,ω, and/or L.
vB=
i^:
j^:
2)For the instant of time described, give the ball's acceleration aB with respect to the fixed ground frame Oxyz. Express your answer using the body coordinates i^ and j^.
Please enter the quantity terms of a,v,α,ω, and/or L.
aB=
i^:
j^:
3)What range of angular velocities is the ball in contact with the outer side of the slot - the slot furthest away from the center of rotation?
Please enter the quantity terms of a,v,α,ω, and/or L.
ω>
To answer these questions, we need to analyze the motion of the ball B relative to the rotating plate and then convert it to motion in the fixed ground frame Oxyz.
1) To find the ball's velocity vB with respect to the fixed ground frame Oxyz, we need to consider two components: the velocity due to the rotation of the plate and the velocity of the ball relative to the plate.
- The velocity due to the rotation of the plate is given by ω cross r, where ω is the angular velocity of the plate and r is the position vector from the center of the plate A to the ball B. Since the ball is located on the j^ unit vector at a length of L from the center of the disk, the position vector r can be written as r = L * j^.
The cross product ω cross r can be evaluated as follows:
ω cross r = (ω * j^) cross (L * j^) = ω * L * (j^ cross j^) = 0
Therefore, the velocity due to the rotation of the plate is 0.
- The velocity of the ball B relative to the plate is v. Since the ball is moving along the slot in the direction shown, the velocity can be written as v = v * i^.
To find the total velocity of the ball B with respect to the fixed ground frame Oxyz, we need to add these two components:
vB = velocity due to rotation of the plate + velocity of the ball relative to the plate
= 0 + v * i^
So, vB = v * i^.
Therefore, the ball's velocity vB with respect to the fixed ground frame Oxyz is:
i^: vB = v
j^: vB = 0 (since the ball is moving along the slot, perpendicular to j^)
2) To find the ball's acceleration aB with respect to the fixed ground frame Oxyz, we follow a similar process as above.
- The acceleration due to the rotation of the plate is given by α cross r, where α is the angular acceleration of the plate. Since the position vector r is perpendicular to the angular acceleration vector α, the cross product α cross r will be 0.
Therefore, the acceleration due to the rotation of the plate is 0.
- The acceleration of the ball B relative to the plate is a. Since the ball is moving along the slot, the acceleration can be written as a = - a * j^ (negative sign since it's in the opposite direction to the positive j^ direction).
To find the total acceleration of the ball B with respect to the fixed ground frame Oxyz, we need to add these two components:
aB = acceleration due to rotation of the plate + acceleration of the ball relative to the plate
= 0 + (- a * j^)
So, aB = - a * j^.
Therefore, the ball's acceleration aB with respect to the fixed ground frame Oxyz is:
i^: aB = 0 (since the ball is moving along the slot, perpendicular to i^)
j^: aB = -a
3) To determine the range of angular velocities at which the ball is in contact with the outer side of the slot (the slot furthest away from the center of rotation), we need to consider the conditions for contact.
- The ball will be in contact with the outer side of the slot when its position is at the maximum distance from the center of rotation (L). This occurs when the ball is at the topmost or bottommost point on the slot.
At these points, the radial component of the velocity (v) should be zero. Since the ball is moving along the slot and the radial component is perpendicular to the slot, the radial component of the velocity is given by v * sin(45°) = v / √2.
Therefore, the range of angular velocities at which the ball is in contact with the outer side of the slot is when the radial component of the velocity is zero:
v / √2 = 0
This gives us v = 0.
Since the angular velocity (ω) is related to the linear velocity (v) by the equation ω = v / R, where R is the radius of the circular plate (L), we can substitute v = 0 into the equation to find the range of angular velocities:
ω = 0 / L = 0
Hence, the range of angular velocities at which the ball is in contact with the outer side of the slot is ω > 0 (greater than zero).