15.13 : 运动方程：一般平面运动-解决问题

Consider a lawn roller of mass 100 kg, a radius of 0.2 meters, and a radius of gyration of 0.15 meters. If a force of 200 N is applied with an angle of 60 degrees with the horizontal, then what is the angular acceleration of the lawn roller?

The coefficient of static friction between the ground and the lawn roller is 0.15, and the coefficient of kinetic friction is 0.1.

Assuming rolling without slipping, the moment of the point of zero instantaneous velocity, point A, is calculated using a horizontal component of the applied force.

At point A, the moment of inertia is calculated using the parallel axis theorem.

Substituting the value of the moment of inertia at point A in the moment equation gives the value of angular acceleration.

The assumption of rolling without slipping motion is valid if the frictional force resulting from the movement of the lawn roller's center is lower than the maximum static frictional force.