Does the centre of mass of body rolling on a stationary floor experience centripetal acceleration downwards towards the IAOR (instantaneous axis of rotation) which here is the point of contact of body with the ground? if no why not? We do it in the case of hinged rod (about corner) falling from horizontal position in the vertical plane. if yes why the com does not move downwards?
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Consider a horizontal floor. A free body diagram of the rolling body (mass $m$) would show that the net force toward the IAOR is $mg-\mathcal{N}$ where $\mathcal{N}$ is the normal force of the floor on the body. The acceleration toward the IAOR is then $$a = g-\frac{\mathcal{N}}{m}.$$
One could consider this to be a centripetal-type of acceleration and get $$ g-\frac{\mathcal{N}}{m} = \frac{v^2}{R}$$ where $R$ is the radius of curvature of the floor. If the floor is horizontal, the radius of curvature would be infinite ($R\to \infty$), then the normal force would be equal in magnitude to $mg$, which is what is expected. If the floor has finite curvature, then the normal force would vary with the curvature and the velocity. The velocity will change rolling on a curved floor, but I'll let you figure out how that is related to floor curvature. An important thing to realize is the the normal force is not always $mg$.
