Suppose a sphere is rolling down an inclined plane. There is no friction. The body will not roll and undergo just a translation. But why is this so? If we consider the axis to be along the point of contact, then there would be a torque which will cause it to rotate but in reality the body won't rotate. Why is this so?
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asked Jun 9, 2016 at 4:31
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$\begingroup$ I believe that it has to do with the fact that a body always prefers to have lowest possible energy. Since in translation, the kinetic energy is less than that of the same body undergoing rolling , therefore it prefers translation more than rolling. In a case of no friction, there is no agent to avoid the slipping of the body. And basically I think its a case of Slipping .@JohnRennie mentioned in a comment that there will be a Force parallel to the plane. Yes! There would be and would provide torque too to cause rotation. But to observe that rotation, we might need a very long incline $\endgroup$– JdeepMay 27, 2020 at 7:09
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$\begingroup$ Also , in rolling ,the bottom most point (point of contact) has 0 velocity. Why does it has 0 velocity? Because of friction!!! . It is friction that stops the relative motion between two particles. Without friction, there is no agent to stop the relative motion between the bottom most point (of the body) and the surface. Hence the velocity of the top most point is same as velocity of the bottom most point . Thus the body undergoes translational motion $\endgroup$– JdeepMay 27, 2020 at 7:14
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3 Answers
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In such a hypothetical situation in which there is no friction between the sphere and plane, there can be no tangential force acting on the sphere, and hence no torque. The only force acting on the sphere would therefore be its weight, and the component of that force acting perpendicularly to the plane would be responsible for its translation down the plane, without rolling.
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1$\begingroup$ Is that true? Wouldn't the centre of mass of the sphere be downhill of its contact with the plane, in which case that would create a torque. $\endgroup$ Jun 9, 2016 at 5:15
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2$\begingroup$ What a wonderful question this is! Yes, the centre of mass would be downhill of its contact. However, in the absence of friction, the only forces acting on the sphere would be its weight and the normal force acting perpendicularly to the plane. Both would pass through the centre of the sphere and hence cannot create a torque. $\endgroup$– PODJun 9, 2016 at 5:25
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$\begingroup$ Ah yes, of course, there can be no component of the force parallel to the plane $\endgroup$ Jun 9, 2016 at 5:55
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1$\begingroup$ I think that the question is about the net torque about the point of contact between the sphere and the slope? $\endgroup$– FarcherJun 9, 2016 at 6:27
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The question answered by @SatwikPasani is related but not quite a duplicate.
The apparent paradox is resolved by realising that using a frame of reference relative to the sphere which is accelerating down the slope is a non-inertial frame of reference.
If there is friction and the no slipping condition is satisfied then the frame of reference attached to the point of contact $P$ is an inertial frame because momentarily the sphere at that point of contact is not moving relative to the Earth which is assumed in such examples to be a good approximation to an inertial frame.
When there is no friction the point of contact on the sphere $P$ is accelerating relative to the Earth and so a pseudo force $mg \sin \theta$ up the slope (green) and passing through the centre of mass of the sphere $C$ has to be introduced if Newton's laws are to be used in that non-inertial frame.
Remember that when you sit in that frame of reference the sphere is stationary relative to you and so the net force and torque on the sphere as you see it must be zero.
So with the introduction of pseudo force, the net force and the net torque acting on the sphere are zero.
If you take a stationary point $P$ on the slope as defining your frame of reference then there is a torque about $P$ on the sphere due to the component of the weight down the slope $mg \sin \theta$ and that torque causes the centre of mass of the sphere $C$ to have an angular acceleration about point $P$.
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Consider to figure below (there is no friction):
If we calculate the net torque about point $\textrm O$, we will have: $$\Sigma M_{\textrm O}=-FL$$ Can we say that the block will rotate about point $\textrm O$? Yes, we can. But, this rotating isn't equal to rolling. This rotation is a particle rotation about a point. Because particles of the sphere don't move relative to each other. They move simultaneously on parallel paths. When we say sphere is rolling, we mean that sphere is rotating as a rigid body not a particle.
Equations of motion in plane for a rigid body are expressed below: $$\Sigma \vec F=m\vec a_{\textrm G}$$ $$\Sigma M_{\textrm G}=I_{\textrm G}\alpha$$ $\textrm G$ is the center of the mass.
So, for a rigid body, the necessary condition for rotating is $\Sigma M_{\textrm G}\neq 0$
P.S. Thanks to @garyp for reminding:-)
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$\begingroup$ Correct (depending on your definition of rotation), but there will be angular acceleration about O. But what I'd really like clarified is how this resolves the paradox. $\endgroup$– garypJun 9, 2016 at 12:25
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$\begingroup$ @garyp I say there is no paradox for resolving. $\endgroup$– lucasJun 9, 2016 at 12:35
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$\begingroup$ @garyp "there will be angular acceleration about O" If so, then the sphere in the question will have an angular acceleration too. There is no paradox again. $\endgroup$– lucasJun 9, 2016 at 12:44