What minimum banking angle is required for an olympic bobsled to negotiate a 100-m radius turn at 35 m/s without skidding? (ignore friction.) ?

51 degrees. Since we're ignoring friction, we have to have a banking angle such that the normal force is exactly perpendicular to the banked curve. Since this problem says "ignore friction", if the bank angle is too shallow, the bobsled would slide outwards if the banking angle is too shallow and would fall inwards if the banking angle is too steep. So we have to exactly match the calculated centripetal acceleration. The equation for centripetal acceleration is: F = mv^2/r I'll assume a mass of 1 kg to keep the math simple. Any mass could be used and the direction vectors would be the same except their magnitude would differ. So F = 1 kg * (35 m/s) ^2/100 m F = 1225 kg*m^2/s^2 / 100 m F = 12.25 kg*m/s^2 The local gravitational acceleration is 9.8 m/s^2, so the sum of those vectors will have a length of sqrt (12.25^2 + 9.8^2) and an angle of atan (9.8/12.25) below the horizon. The magnitude of the vector doesn't matter, merely the angle which is: atan (9.8/12.25) = atan (0.8) = 38.65980825 degrees. The banking angle needs to be perpendicular to the force vectors. So 90 - 38.65980825 = 51.34019175 degrees. Rounding to 2 significant figures gives a bank angle of 51 degrees.