- #1

- 9

- 0

- I
- Thread starter WilkinzMicawber
- Start date

- #1

- 9

- 0

- #2

- 950

- 418

x = sqrt(m/k) v0

The force from the first contact to maximum compression is -k x, so the acceleration is

a = - kx/m.

dx^2/dt^2 = -kx/m

Implies

x = C sin(sqrt(k/m) t)

Which knowing the max x from above must be

x = sqrt(m/k) v0 sin( sqrt(k/m) t)

where t=0 gives x=0 as the moment of initial contact. Maximum compression is reached when

sqrt(k/m) t = pi / 2

So tmax = sqrt(k/m) pi/2

So higher initial velocity gives greater compression but the time of contact is the same.

Regarding the effect that has on the transfer of the angular momentum for some of the time the ball slips and there is one torque then it sticks. Since the force is higher the torque is higher but once it's enough to stop the ball I suppose there isn't much difference if you go to yet higher velocity. I have no idea at what velocity the spin finally stops.

- #3

- 9

- 0

- Replies
- 7

- Views
- 5K

- Replies
- 1

- Views
- 12K

- Replies
- 3

- Views
- 7K

- Replies
- 7

- Views
- 731

- Last Post

- Replies
- 3

- Views
- 3K

- Replies
- 2

- Views
- 733

- Replies
- 32

- Views
- 5K

- Replies
- 7

- Views
- 3K

- Replies
- 10

- Views
- 13K

- Replies
- 16

- Views
- 7K