Question

4) Tiger O’Connor is preparing for his new career on the PGA tour. He wants to increase his driving distance by lengthening his driver. He decides to build a new driver that identical to his old driver except for the fact that the new driver is six inches longer. The masses are the same (0.25 kg) and the moment of inertia about the center of mass for both drivers is 0.05 kg · m2 . The distance from the axis of rotation to the center of masses are 0.97 m and 1.14 m. a) What are the moments of inertia of each driver about the axis of rotation? (10 points) b) If the same torque is applied to the swing of each driver (15 Nm), what are the angular accelerations of each club? (10 points) c) What is the angular velocity of each club at the point of contact? Assume that both swings move through the same range of motion from the top of the swing to contact (θ = 3π/2). Hint, use an angular variation of one of the equations of linearly accelerated motion (ωf 2 = ωi 2 + 2αθ). (10 points) d) Which driver will results in a greater club head speed? The length of the drivers are 1.08 m and 1.25 m. (10 points)

Answer 1

a) we use parallel axis theorem

Moment of inertia of old driver about axis of rotation = MoI about Centre of mass + Md^2

   = 0.05 + 0.25*0.97*0.97 = 0.05 + 0.235 = 0.285 kg m2

Moment of inertia of new driver about axis of rotation = MoI about Centre of mass + Md^2

   = 0.05 + 0.25*1.14*1.14 = 0.05 + 0.325 =0.375 kg m2

b) angular acceleration of old driver = torque/MoI = 15/0.285 = 52.63 rad s-2

angular acceleration of new driver = torque/MoI = 15/0.375 = 40 rad s-2

c) angular velocity of old driver = (2*angular acc.*angular displacement)^1/2 = (2*52.63*3/2)^1/2 = 22.27 rad s-1

angular velocity of new driver = (2*angular acc.*angular displacement)^1/2 = (2*40*3/2)^1/2 = 19.41 rad s-1

d) club head speed of old driver = angular velocity * length = 22.27*1.08 = 24.05 m s-1

club head speed of old driver = angular velocity * length = 19.41*1.25 = 24.26 m s-1