A lot of people have measured how fast Federer hits the ball and how many rpm he can generate, but nobody has ever measured racket speed. Using basic physics, it is fairly easy to theoretically calculate what nobody has experimentally measured.
The force Federer exerts on the ball is related to Impulse, or change in momentum. Specifically,
Impulse = Average Force × Δt = Δp
where Δp is equal to the change in momentum Δt is equal to the time interval during which the change of momentum occurs. Change in momentum can be described as,
Δp = final momentum - initial momentum = m2v2 - m1v1
Lets say that Roger Federer is playing Rafael Nadal,and Nadal just hit a defensive topspin forehand at about 50 mph. There is enough topspin on the ball to maintain the ball's speed after it bounces, so the moment before Federer swings the ball is still moving at 50 mph. After the collision, the ball is moving back at Nadal at about 80 mph, the average pace of Federer's forehand. If the ball is in contact with the racket for 16 ms (see study of ball-and-racket contact times) and weighs about 57 grams, then the average force Federer is exerting on the ball is
Average Force = Δp/Δt = (m2v2-m1v1)/Δt = [(.057kg)(35.8m/s+22.3m/s)]/.016s = 206 N
If Federer were swinging his racket horizontally, then 206 N would be the total force he was exerting on the ball. Unfortunately, this is not the case. Watch the following video, and see how Federer is swinging his racket at an upward angle as he hits the ball.
The force calculated above is only one component of the total force Federer is exerting on the ball. He is also exerting an upward force that contributes to the total force with which Federer swings his racket.
In the video, it seems as though Federer is swinging upwards at about a 30° angle. Basic component addition could give us the total force with which Federer is swinging his racket.
cos30° = Fx/Ft
Ft = Fx/cos30° = 206 N/cos30°
Ft = 239 Newtons
Now that we know the approximate force with which Federer hits the ball, how do we convert it to speed. Lets begin with calculating the torque acting on Federer's arm. Torque is defined as rF, where r is radius and F is force. We already know what the force is - what is the radius? Roger Federer is about 1.85 meters tall, so assuming that arm length is about about 3/8 of total height, his arm is about .7 meters long. His racket probably adds another .5 meters to the total length, so the total radius of the swinging arm should be approximately 1.2 meters. The actual length is probably a little smaller than this, because Federer's arm is not fully extended while he is swinging (it is slightly bent). According to these numbers, torque is approximately;
τ = rF = (1.2m)(239 N) = 286.8 N·m
As we all know, torque is also defined as Iα, where I is the moment of inertia and α is the angular acceleration. Assuming that Federer's arm acts like a rod rotating around one end,
I = (1/3)ML^2
According to various human studies, the mass of an arm is approximately 5% of total body weight. Federer weighs 88 kg, so his arm should be approximately 4.4 kg. His racket adds an extra .363 kg, so total mass of the swinging arm is approximately 4.76 kg. Using all this information, it is possible to calculate the approximate angular acceleration of Federer's racket.
τ = Iα
α = τ/I = τ/[(1/3)ML^2]
α = 286.8/[(1/3)(4.76)(1.2^2)]
α = 125.5 rad/s^2
The last step in this calculation is to convert the angular acceleration to an angular velocity. Rotational kinematics should make this a fairly simple process. Assuming that Federer starts his swing with zero angular velocity, I can use the equation ω^2 = 2α(Δθ). The last variable needed to solve this equation is Δθ - the change in angular displacement. To determine this value, I have drawn an aerial diagram of how Roger Federer swings the racket. This, like most real-world calculations, is an approximation and should not be taken as the true value. In the diagram, Federer begins his swing with the racket fully behind him and hits the ball as it is passing him. The angular displacement in this diagram is approximately π/2 radians. Plugging these numbers into the kinematic equation, ω is approximately,
ω^2 = 2α(Δθ)
ω = [2(125.5 rad/s)(π/2)]^1/2
ω = 19.9 rad/s
According to these calculations, Federer is swinging his racket with an angular velocity of approximately 19.9 m/s. In terms of linear velocity, Federer is swinging his racket at about 24 m/s (ωr). I don't know about you, but accelerating a tennis racket that weighs 1/3 of a kilo to a speed of 24 m/s in a few feet is pretty impressive to me!