There and back again
Suppose you want to throw a super ball so that it bounces back and forth, retracing the same trajectory, as shown above. If you throw the ball horizontally with velocity
1
m
/
s
,
what spin
ω
(in
rad
/
s
) must you give it so that it does what you want?
Assumptions and Details
- The collision with the ground is perfectly elastic.
- The ball does not slip while it contacts the ground.
- The radius of the ball is 3 c m .
The answer is 83.333333333.
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2 solutions
:D I solved it almost exactly like this
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I found this problem to be a nice one! =D
I just have a question... Doesn't the question states "The collision is perfectly elastic" means that "There are no friction between them"? Because of this, I thought there isn't any horizontal impulse act on the ball..
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Hi Kelvin. Perfectly elastic collisions don't mean no friction between colliding surfaces. They just mean conservation of kinetic energy, which doesn't depend on direction of motion. The impulse on the ball can be in any direction as long as kinetic energy is conserved. In order to follow the path described, it should be such as explained in m solution.
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Oh, now I know, so the friction force doesn't always relevant to the change of kinetic energy. Thanks for explanation.
Answer is ω = 2 R 5 V x
This is very interesting situation , and we can understand's many thing's from this , Like what is minimum and maximum coffecient of friction ? , what should be nature of ball, means what type of material is used in making such Super ball? ...etc.etc.
So Check out This discussion which i posted a while ago :
The link you mentioned is UNDEFINED, much like y'(0) of y=|x| >.< :P -_-
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Thanks ! I had edited that , you may check it now .
I too used that! :D
I was thinking about the situation in which ball retraces it's path only after 1st collision not after successive collision. In such case ω f = − ω i .
f j = 2 m V x
f R = I ( ω f − ω i )
As the ball is not slipping so just after the collision v = r ω
So V x = R ω f .
Are these correct??
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How do you mean
I was thinking about the situation in which ball retraces it's path only after 1st collision not after successive collision.
If it retraced its path the first time, would it not have the same initial conditions going into the next collision?
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In order to retrace it's path the direction horizontal velocity of the ball after collision must be opposite to the direction it's horizontal velocity before collision. It order to do it multiple times the sign of ω must be changed after each collision to satisfy the above condition.
But if it has to retrace it's path only after 1st collision then direction of horizontal velocity must be changed after 1st collision but now there are no constraint on angular velocity. So in this condition ω f = ω i .
It's just a confusion. Sorry, if I am wrong. :)
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@Satvik Pandey – You are right, it can bounce back with the same horizontal speed (so it will trace out the same path) but have a different angular speed (so the conditions will be different in the next collision).
What I don't understand is your equation v=rw .... I see no reason why it should be true.
When it collides there will be some horizontal impulse J x = m Δ V x which will also provide a "torque-impulse" (sorry for making up words haha) equal to R J x = I Δ ω
Thus we can say: m R Δ V x = I Δ ω
I don't believe we can say much more than this about the motion. The exact details would depend on other information like the elasticity of the collision.
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@Nathanael Case – According to question the ball doesn't slip when it in contact with the ground. The horizontal impulse acting on the ball changes it's velocity of CoM and impulsive torque changes the velocity of the ball about it's CoM. But during time period the ball doesn't slips. So at every instant V x = R ω must be satisfied.
What do you think?
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@Satvik Pandey – Ah, I see. I agree with you (but I may be wrong too of course).
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@Nathanael Case – Now I have one more doubt. :D
According to answer, in order to retrace it's path the angular velocity of the ball should be 8 3 . 3 . But if the ball has to roll in the time when it is contact with the ground then ω f = V x / R = 1 0 0 / 3 .
So, I think that ball will not roll in time when it is contact with the ground (in this case).
What do you think?
@Nathanael Case @Deepanshu Gupta
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@Satvik Pandey – I think the constraint v x = ω R does not need to be true because the ball will be deformed, so it is not the same a sphere rolling.
This paper does not directly answer you, but it might be of interest: Garwin Super Ball
When the super ball is about to collide with the floor, it has an horizontal velocity v = 1 m/s and an angular velocity ω . Just after the collision, in order for the ball to bounce back and forth retracing the same trajectory, the horizontal and angular velocities of the ball should be − v and − ω , respectively. Hence, the change in linear and angular momentum is Δ p = − 2 m v and Δ L = − 2 I ω , respectively, where m and I = 2 m R 2 / 5 are the mass and the moment of inertia of the ball and R = 3 cm is its radius.
There is an horizontal force, say F that produces such changes for a small period of time, say Δ t , such that
F = Δ t Δ p = Δ t − 2 m v
F R = Δ t Δ L = Δ t − 2 I ω
Replacing the first one into the second one we get
ω = 2 R 5 v
which replacing numeric values we get ω = 8 3 . 3 3 rad/s.