+10 votes
in Physics by
edited by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
+7 votes
by

When an object moves in a circular path, it experiences a centripetal force directed towards the center of the circle. Friction can indeed provide this necessary centripetal force, even though it acts in a direction opposite to the object's velocity.

Let's consider an example of an object moving in a horizontal circle on a rough surface. As the object moves, it exerts a tangential force on the surface due to its inertia. Simultaneously, the rough surface exerts an equal and opposite force on the object due to friction, according to Newton's third law.

Now, the crucial aspect to understand is that the frictional force is not solely responsible for providing the centripetal force. Instead, it is the horizontal component of the frictional force that acts towards the center of the circle, contributing to the centripetal force.

When an object moves in a circular path, its velocity vector is tangent to the circle at any given point. The frictional force, being parallel to the surface, has both a vertical and a horizontal component. The vertical component counteracts gravity or any other vertical forces, while the horizontal component provides the necessary inward force towards the center of the circle.

By adjusting the magnitude of the frictional force, it is possible to match the required centripetal force for the object to continue moving in the circular path. If the frictional force is insufficient, the object would move in a larger radius circle or slide outwards. On the other hand, if the frictional force is excessive, it can cause the object to move in a smaller radius circle or even experience slipping.

In summary, although the frictional force acts opposite to the object's velocity, its horizontal component can provide the necessary centripetal force by acting towards the center of the circular path. This allows the object to maintain its motion along the desired circular trajectory.

Welcome to Physicsgurus Q&A, where you can ask questions and receive answers from other members of the community.
...