Question

You have been asked to design a new stunt for the opening of an ice show at Baxter Arena. A small 50 kg skater glides down a ramp and along a short level stretch of ice. The skater then grabs the bottom end of a large 180 kg vertical rod which is free to turn vertically about an axis through its center. The plan is for her to hold onto the 6 meter long rod while it swings her 180° up to the top. The rod has a uniform mass distribution. You have been asked to give the minimum height of the ramp.

I also need a physics descritption of the problem with all the equations laid out. Thank you

Answer 1

conservation of energy between the top and bottom of the ramp is

                                                   E1 = E2

                                         mgh = 1/2*mv^2

                                                      v = sqrt(2gh)

conservation of angular momentum just before and after impact with the rod

                                                  L2 = L3

                                            mv(d/2) = I*w

      m*sqrt(2gh)(d/2) = I*w

                                                    w = (m)(d)(sqrt(2gh)) / (2*I)

                 conservation of energy between 3 and 4

                                              E3 = E4

                                     1/2*Iw^2   = m*g*d

            1/2*I ((m^2*d^2*2gh)) / 4I^2 = m*g*d

                                                  h = 4*I / m*d

I_rod + I_skater = 1/12*Md^2 + 1/4*md^2

                     h = 4/md*(1/12*Md^2 + 1/4*md^2)

                     h = ( 1/3 (M/m) + 1)*d

                        = ( 1/3 (180/50) + 1)* 6m

                     h = 13.2 m

the minimum height of the ramp h = 13.2 m