Dancers experience large forces associated with the jumps they make. for example, when a dancer lands after a vertical jump, the force exerted on the head by the neck must exceed the heads weight by enough to cause the head to slow down and come to rest. the head is about 9.4% of a typical persons mass. video analysis of a 65-kg dancer landing after a vertical jump shows that her head decelerates from 4.0 m/s to rest in a time of 0.20 s.

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Emelia Lang

Physics

12 July, 05:54

Dancers experience large forces associated with the jumps they make. for example, when a dancer lands after a vertical jump, the force exerted on the head by the neck must exceed the heads weight by enough to cause the head to slow down and come to rest. the head is about 9.4% of a typical persons mass. video analysis of a 65-kg dancer landing after a vertical jump shows that her head decelerates from 4.0 m/s to rest in a time of 0.20 s.

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Grady Costa · Answer 1

So the person weighs 65kg, and 9.4% of that is head, so 65*0.094 = 6.11 kg is the mass of the head. If we deaccelerate from 40 m/s to 0 m/s in 0.2 s, our total acceleration is: a = Δv/Δt = (0 - 40) / (0.2 - 0) = - 200 m/s². We can then use Newton's second law, F = ma, to find the force, using m as mass of the head and a as our acceleration (we'll ignore the negative sign because we don't care about the force's direction here). F = ma = (6.11) (200) = 1222 N, a pretty large amount of force.

It now occurs to me that the easier way to do this, though slightly more advanced, is to use that Force is the derivative of momentum, or F = dp/dt, or with no calculus, F = Δp/Δt, where p is momentum and t is time. p = mv, where m is mass and v is velocity, so F = Δp/Δt = Δ (mv) / Δt = ((6.11) (0) - (6.11) (40)) / (0.2 - 0) = (6.11) (40) / (0.2) = 1222 N. So yeah it's quicker, I feel this is less straight forward though.