Question

The roller coaster car has a mass of 850 kg including its passenger. It is released from rest at the top of the hill A Neglect friction, the mass of the wheels, and the size of the car. Take h_B = 19 m, rho_B = 9.5 m. h_C = 12 m and rho_c = 6 m. Part A Determine the minimum height h of the hill crest so that the car travels around both inside the loops without leaving the track. Part B What is the normal reaction on the car when the it is at B? Part C What is the normal reaction on the car when the it is at C?

Answer 1

According to conservation of energy Rightarrow k E_A + PE_A = KE_B + PE_B since friction is neglected Rightarrow 0 + 0 = 1/2 (850 k g) (V^2B) - 850 kg times 9.8 (h - 19 m) Rightarrow 1/2 (850 kg) (V^2_B) = 850 kg times 9.8 (h - 19 m) straightline 1 and gravity force F_B = mg = mV^2_B/r Rightarrow (850 kg) (9.8) = (850 kg) V^2_B/9.5 m Rightarrow V_B = 9.65 m/s plug in equation 1 Rightarrow 1/2 (850 kg)(9.65)^2 = (850 kg)(9.8) (h - 19 m) Rightarrow h = 23.75 m the minimum height of the hill is 23.75 m at B: - Rightarrow N_B = 0 (Since h to be minimum) \r\n at C: - Rightarrow KE_A + i E_A = K E_c + pE_c since friction is neglcted Rightarrow 0 + 0 = 1/ 2 (850 kg) (V^2_c) - (850 kg) times 9.8 (23.75 - 12) Rightarrow 1/2 (850 kg) (V^2_c) = (850 kg)(9.8) (11.75) V_c = 15.175 m/s Net force: Rightarrow N_c + mg = m v^2/r