# Doubt related to rotational motion (rolling down an inclined plane)

I have doubt regarding when an object ( sphere , ring , cylinder etc ) rolls down a FRICTIONLESS inclined plane.

Let us suppose the angle of inclination of inclined plane is θ.

Now the forces acting on it will be weight force (mg) and normal reaction. A component of mg perpendicular to incline mgcosθ will be balanced by the normal reaction. There will be another component of mg along the incline mgsinθ which will give the object acceleration gsinθ. There is no friction.

Now, acceleration, a = radius ( r ) x angular acceleration (α) Now as there is, a = gsinθ so there must be nonzero value of α.

But force mg and normal reaction are forces whose line of action passes through the centre of the object ( ring etc ) ( passes through centre of mass). So net torque about centre should be zero and there should be no angular acceleration.

Also, will the angular momentum be conserved when it reaches the bottom of incline?

• Don't use capital letters to reflect emphasis; it rather increases the vagueness. Either italicise or make them bold . Also this is a mathjax-enabled site which supports $\LaTeX$. Try to use them. – user36790 Oct 9 '15 at 17:28

• @user21540 With all of that said, if my teacher/professor asked me to work out a problem involving a ball rolling down a frictionless ramp, I would understand him/her to most likely mean that the rolling friction coefficient was zero but that the ball was indeed rolling without slipping, hence $\frac 12m|\vec v|^2 + \frac12 I|\vec v/R|^2 - m\vec g\cdot\vec r = \text{constant}$, where $\vec r$ is the position of the center-of-mass, R is the radius of the ball, $\vec g$ is the gravitational acceleration, and $\vec v = \frac{d\vec r}{dt}.$ – CR Drost Oct 9 '15 at 20:42