Force exerted by a rolling body

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When a block of mass m slides down a frictionless inclined plane at angle ∅, it exerts a force of mgcos∅ perpendicular to the surface. In contrast, a rolling cylinder of the same mass also exerts a perpendicular force of mgcos∅ on the inclined surface, regardless of its rolling motion. If the surface is frictionless, the cylinder will not roll due to lack of angular acceleration. The discussion clarifies that the normal force remains mgcos∅ for both cases, as there is no additional force component acting perpendicular to the surface. Thus, the force exerted by the rolling cylinder does not exceed mgcos∅.
sodaboy7
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Lets consider a block of mass m, sliding down an inclined surface (no friction) at an angle ∅ with ground. The force exerted by this block on the inclined surface is mgcos∅ which is perpendicular to it. Now consider a cylinder of same mass m and rolling (without slipping) down the same plane. What will be the force exerted by the rolling cylinder on inclined surface? will it be mgcos∅ or something more than that due to its rolling motion.
 
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If the surface is frictionless, the cylinder isn't rolling, since it will have linear acceleration, but not angular acceleration. If there's friction, then that will produce an upwards force in the sin(θ) direction.
 
in case of block its frictionless and in case of rolling its not. I am asking about forces perpendicular to surface not along the surface. So will there be any force other than mgcos∅?
 
The normal force will be the same: mgcos∅
 
sodaboy7 said:
I am asking about forces perpendicular to surface not along the surface. So will there be any force other than m g cos(θ)?
Since's there's no component of acceleration in the direction of cos(θ), the component of force from gravity and the opposing normal force from the plane in the direction cos(θ) will be m g cos(θ) (as Doc Al just posted), in either the sliding or the rolling case.
 
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