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>
Redraw the Axy coordinates system on a piece of paper, draw x horizontally and y vertically. Place the ball B on your drawing. x position is easy to find, y is a function of 45deg and L.
To find vrel:
Look at the drawing in the exercise and draw the same vector v on your drawing. Ask yourself a question: in what direction does the ball move? Has it any velocity in x direction? In y direction?
To find arel:
Do the same.
to find vB:
Find the center of rotation of the disk. Take the distance between the center you've just found and the center of the ball. Ask yourself a question: How to find a linear velocity of a point on the rotating disk (it's a function of omega and L)? Now you've 2 parallel vectors. Does they point in the same or opposite direction (one vector is known, to find the direction of the other one use right hand rule)? Do you have any velocity in y direction?
to find aB:
take the vB and differentiate it: ask yourself a question, which factor changes in time? Use chain rule. Don't forget about vectors i and j. Do they change in time? What do they give when differentiated? (see Derivative of a vector in a rotating frame). There's a catch: if you have in your result something like omega*v*j write it as 2*omega*v*j (Coriolis acceleration).
its vB L*omega/2
To answer these questions, let's break down the given information and solve step by step:
1) To find the ball's velocity with respect to the fixed ground frame Oxyz, we need to consider the velocities of both the rotating plate and the ball. The velocity of the rotating plate can be found using the equation:
vA = r x ω
Here, r is the vector from the center of the plate to the point of contact with the ball, which is Lj^ (since the ball is located on the j^ unit vector at a distance of L from the center), and ω is the angular velocity of the plate. Since the plate is rotating along the i^ axis, the cross product r x ω will have a j^ component equal to r * ω, where * denotes the cross product in 2D.
vA = L * ω
Now, the velocity of the ball with respect to the plate is given as v.
Therefore, the velocity of the ball with respect to the fixed ground frame Oxyz, vB, is the vector sum of vA and v:
vB = v + vA
Since v is not given in the question, we cannot determine the exact values of vB. However, we can express the answer in terms of the given quantities by substituting vA = L * ω:
vB =
i^: v (component along i^ direction)
j^: v + L * ω (component along j^ direction)
2) To find the ball's acceleration with respect to the fixed ground frame Oxyz, we use a similar approach. The acceleration of the rotating plate can be found using the equation:
aA = r x α + r' x ω
Here, r' is the vector from the center of the plate to the point of contact with the ball (which is the same as r), and α is the angular acceleration of the plate. As before, the cross products will have the same j^ component, so:
aA = L * α + L * ω
Now, the acceleration of the ball with respect to the plate is given as a.
Therefore, the acceleration of the ball with respect to the fixed ground frame Oxyz, aB, is the vector sum of aA and a:
aB = a + aA
Again, since a is not given in the question, we cannot determine the exact values of aB. However, we can express the answer in terms of the given quantities by substituting aA = L * α + L * ω:
aB =
i^: a (component along i^ direction)
j^: a + L * α + L * ω (component along j^ direction)
3) To find the range of angular velocities where the ball is in contact with the outer side of the slot (furthest away from the center of rotation), we need to consider the conditions for contact. As mentioned in the question, the ball is slightly smaller than the slot, so it only contacts one side of the slot at a time. This implies that the ball will be in contact with the outer side of the slot if its position along the slot is greater than L.
The position of the ball along the j^ direction is given by the vector AB, which has a magnitude of L. Since the ball is located at the end of this vector, the range of angular velocities where the ball is in contact with the outer side of the slot is:
ω > 0 (clockwise rotation) and ω < π/L (anti-clockwise rotation)
Note that the range is given in terms of angular velocities, not specific values.
I hope this helps! Let me know if you have any further questions.