
#1
Nov1505, 05:32 PM

P: 48

Wich is the best way to estimate contacttime between racquet and a tennisball, under assumption that the force acting is constant? Also known is the speed of the tennisball after the stroke, v=27.8 m/s (100km/t), and the dimension of the ball under the stroke is half of normal.




#2
Nov1505, 08:28 PM

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P: 1,322

The force of the racket on the tennis ball has the effect of accelerating the tennis ball's center of mass. If the radius of the tennis ball is R, then the center of mass of the tennis ball is initially a distance R/2 from the racket (the dimension is half according to your description). Please tell me if I have interpreted something wrong so far. What are your given quantities?
The contact time for this phase of the stroke will then be the time it takes for the center of mass to move from R/2 to R, why? Can you find this time in terms of given quantities using your knowledge of the final velocity? 



#3
Nov1505, 09:05 PM

P: 48

Using the assumption that the center of mass under the stroke will move 2R i got the result t=5.8ms. this sounds fair, i guess..
found it using, m=0.1kg R=4cm and: F=Ekin/2R and then t=mv/F=5.8ms 



#4
Nov1505, 09:31 PM

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Estimating contacttime
How did you decide that the center of mass would move 2R? You need to be careful about what you assume the initial condition is. I described one possible phase of the stroke above, but you need to ask yourself, based on the initial conditions, just exactly what the full stroke is.




#5
Nov1505, 11:30 PM

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If you assume that the force is constant, F, for a time [itex]\Delta t[/itex] the impulse [itex]F\Delta t[/itex] is equal to the momentum of the ball. Also, [itex]Fd/2 = mv^2/2[/itex] where d is the diameter of the ball. Combine the two equations to get rid of m to find the time and you end up with: [tex]mv/\Delta t = mv^2/d[/tex] [tex]\Delta t = d/v[/tex] But I don't think it is a reasonable assumption. The racquet is moving and the racquet strings contract as the ball remains in contact with the racquet. You have to take into account the distance that the force is applied through in the original frame of reference (of the stationary tennis court). This could be considerably larger than half the diameter of the ball. AM 



#6
Nov1505, 11:41 PM

P: 543

I think all you need to do is , considering absence of any external forces , the Force applied over time 't' would give impulse tot he ball initially assumed to be at rest.
So F.t = Change in momentum This would give you the answer including the compression effects , which would lead to only internal forces. BJ 



#7
Nov1605, 08:03 AM

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AM 



#8
Nov1605, 03:19 PM

P: 48

another thing i wondered about was:
using the following two eq.s ( [itex] F \Delta x = 1/2 mv^2 [/itex] and [itex] F \Delta t = mv [/itex]) you could solve and find [itex] \Delta t = 2 \Delta x /v [/itex]. why is this result different (by factor 2) than the well known [itex] x = x_0+v t [/itex]? and wich is the "right" to use? 



#9
Nov1605, 03:25 PM

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Remember the velocity in [tex] \Delta t = 2 \Delta x / v_f [/tex] is the final velocity, while the velocity in [tex] \Delta x = v_a \Delta t [/tex] is the average velocity. What is the relationship between [tex] v_f [/tex] and [tex] v_a [/tex] for a constant force (assuming the system starts at rest)?



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