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Name: Alicia
Status: student
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If a ball is placed (from rest) on a completely frictionless ramp and released, will it roll or slide down the ramp? I believe the ball will slide rather than roll, but I get confused when I think about torque. Let us put the pivot at the point of contact between the ball and the ramp. Doesn't the weight of the ball (assumed to be acting at the ball's center of gravity) create an unbalanced torque? The normal force from the ramp would act in line with the pivot, and thus would not create a torque. Thanks for your help.

Hello Alicia,

This is a really great question that truly tests a person's understanding of linear and rolling motion. Your intuition is spot on.

If the contact between the surface and the sphere is completely free of friction it will slide without rolling. There needs to be some form of frictional force at the point of contact for rolling to occur. I think your confusion over torque is introduced with the concept of "pivot". Without friction there can be no pivotal point about which to rotate. The point of contact will merely slide.

However, it is perhaps best to actually draw a "free-body-diagram" (I hope that is a familiar word) for the problem. If you put all the forces down from the points about which they act, there will not be any torque. The gravitational attraction will act through the center of the body. The normal force also acts through the center. The decomposition of the forces into their respective X and Y components also act through the center. The only way to get the ball to roll is to add a frictional force at the point of contact that will act to pull the sphere around and cause rolling.

I hope this helps,




It is true that the weight of the ball will create an unbalanced torque around the point of contact between the ball and the ramp. However, the ball will obtain an increase of angular velocity and angular momentum around this pivot point without rolling. The center of mass of the ball is moving perpendicular to the radius vector (from pivot to center of mass). This is angular velocity around the pivot. If the ball were to move in a straight line down the ramp without speeding up or rolling, the center of mass would still have an angular velocity around the point of contact. IF you were a camera at the point of contact, you would have to keep rotating in order to keep pointed at the ball's center of mass.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College


The ball is the inertial object with motions we want to understand. The ramp and gravity are just props, only half-"real", oversimplified and fixed. So start your thought with the ball in free space, no forces acting on it. It has it's center of gravity, which is also its center of rotation and of angular inertia. To change its rate of rotation, some tangential force must be applied to its skin.

Then you apply gravity and the ramp to our ball. Given a frictionless ramp surface and a perfectly spherical ball with center of gravity (CG) at dead center, no such torque can be applied and the ball will only slide, not rotate. The force vector which the ramp applies (normal to its surface) is also radial to the sphere, and therefore pointed directly at the CG. Torque is equal to force times distance, specifically the lateral distance by which the force vector line misses the CG. That distance is zero, so the torque must be zero.

Gravity applies a vertically downward force vector to the CG. It can be resolved into any two orthogonal components. The component of it which is normal to the ramp is opposed by the ramp developing an equal and opposite response force. The remaining component, parallel to the ramp, accelerates the ball parallel to the ramp surface. It too acts on a line directly through the CG and develops no torque.

What changes if a lightly inflated ball develops a broad flat contact spot when pressed against the ramp? Nothing, I think. The array of force vector lines from the area of the spot would be evenly balanced, missing the CG equally on all sides. Net torque would still be zero, and the ball would still slide without rotating.

If the CG were fixed off-center in the sphere, the ball might try to rotate briefly, but after rotating some it would develop opposite torques. You can imagine how an off-center weighted ball might rock back and forth on a level surface. On the ramp it would slide down with some degree of that rocking motion superimposed. If the CG-offset at release was directed away from the ramp, the ball might almost be able to roll over completely. For an idealized frictionless ramp, the rocking and the sliding are completely separable problems. That says everything.

Jim Swenson

If it is frictionless, then it would slide. It can only have a normal force, as you say, and that is in line with the contact point, so causes no torque. The lateral component of the normal force would cause the ball to move down the incline.

David Brandt, P.E.

Good question. Since we do not experience truly frictionless surfaces in everyday events, their properties seem a bit weird, to say the least.

It would seem like the ball would roll, but without friction, it would just slide and not rotate. One needs friction to transfer some translational (straight line) kinetic energy to rotational kinetic energy (spinning).

Let us do a thought experiment. Think of a ball on a vertical wall. Here there is no normal force (gravity and normal force are at right angles) hence no friction between ball and wall. If you let the ball fall, there is no reason to think it will rotate; it will just fall down. This is as close as we can get to a purely frictionless surface and the reasoning holds. The ball will not rotate in the absence of friction.

Another way to look at this; if you are going to trip (the ball to start spinning) your foot needs to stop or be caught on something (a version of friction). Without friction, the seeming imbalance of the ball's center of mass is of no consequence.

Bob Avakian
Oklahoma State University - Okmulgee, OK

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