If a force acts horizontally at the top of a rolling ball, causing a nonzero net torque without slipping, Newton's second law, F = ma, can still be applied to the center of mass of the ball.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In the case of a rolling ball, the net force can be divided into two components: the force causing linear acceleration (F) and the force causing rotational acceleration (τ/R), where τ is the torque and R is the radius of the ball.
When the ball rolls without slipping, the linear acceleration (a) and angular acceleration (α) are related by the equation a = αR. This means that the linear acceleration is directly proportional to the angular acceleration and the radius of the ball.
The torque (τ) is related to the angular acceleration (α) by the equation τ = Iα, where I is the moment of inertia of the ball.
Therefore, combining these equations, we have:
F - (τ/R) = m * a - (I/R^2) * α
Since a = αR, we can substitute αR for a:
F - (τ/R) = m * (αR) - (I/R^2) * α
Rearranging the equation, we get:
F = m * α * (R + I/R^2)
So, Newton's second law, F = ma, still holds true for the center of mass of the ball. The equation includes the rotational terms due to the nonzero net torque and the rolling motion without slipping.