Question

017 10.0 points Consider two bicycle riders, A and B. The two riders have equal masses MderMBde and their respective bicycles also have sim- ilar frames, the wheels of the two bicycles have equal masses MWheel and equal radii MframeMame Finally, eheel but different mass distribu- tions: the wheels of bike A have most of their masses at the rims, Wheel A, while the wheels of bike B have their masses 'spread' evenly over the whole wheel area, Wheel B The two cyclists travel at the same speed on level ground. They approach a low hill and decide to coast up instead of hard pedalling At the top of the hill, which of the two bikes will have a larger speed? Assume no friction

Answer 1

Apply the law of conservation of energy. Energy stored in the riders should be equal to the gravitational potential energy. Consider h is the height of the hill, final velocity of the rider is v. mgh = mv2+2101 Here, m is the total mass of bycycle plus rider, g is the acceleration due to gravity, I is the moment of inertia of the two wheels, and is the angular velocity. Substitute for o (this is the relation between linear velocity and angular velocity) in the above equation. mgh='5m + C) mgh = }( m+ 2mgh m + According to the above equation, velocity depends on the moment of inertia, if moment of inertia is higher then velocity will be lower. Now, consider the expression for the moment of inertia. I = MWheel,.2 Here, Mis the value of mass particals and r is there distance from the centre. In case of wheel A, most of the mass is distributed at the rim (the outer edge of the wheel). While for B, mass is evenly distributed. Hence, moment of inertial of B is greater than moment of inertia of A. Hence, according to the expression of velocity, velocity of A wiil be higher than the velocity of B at the top of the hill.