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Question

A horse completes 9 laps around a circular track in 254 s.

a. What is the period?

b. What is the frequency?

c. If the radius of the track is 110 m, what is the velocity of the horse?

d. What is the horse’s centripetal acceleration?

e. The horse has a mass 453.6 kg, what is the centripetal force it will generate around the

turn?

Answer 1

Use Θ= ωt where ω=2π/period. One lap is 2π rad so 9 laps is 18π. To find period let Θ = 18π = 2π • 254/period and solve for period period = 2π•254/18π which gives 254/9 s and period = 28.22s Since frequency is 1/period frequency= 35.4mHz. The velocity of the horse is v=Rω and v= 110• 2π• .0354 = 24.47 m/s The centripetal acceleration is v²/R and it equals 598.78/110 which = 5.44m/s² The centripetal force is F_c =m v²/R which equals 453.6• 5.44 N = 2467.58 N = 2.4676 kN.