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# Translational Velocity

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An object of your choice, of radius R, is given an initial angular velocity about a horizontal axis of w_0 (omega(initial)). The object is then set down on a surface that has a friction coefficient of u and released. What is the translational velocity of the object as it rolls away from the initial point of contact?

https://brainmass.com/physics/velocity/translational-velocity-504721

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4.) An object of your choice, of radius R, is given an initial angular velocity about a horizontal axis of w_0 (omega(initial)). The object is then set down on a surface that has a friction coefficient of µ and released. What is the translational velocity of the object as it rolls away from the initial point of contact?

Let the moment of inertia of the object (sphere or cylinder or a disc) is I. The angular momentum of the object due to rotational motion just before it is placed on the surface about its axis will be

As it is rotating only its angular momentum about any axis parallel to its axis will be the same L0. Thus its angular momentum just before it is placed on the surface about the axis passing to the contact point and normal to the plane of paper is

As the body is placed on the surface, for some time friction will act on it in forward direction reducing its angular velocity and increasing its translational velocity. As the friction as well as weight of the object passes through the point (or line in case of cylinder) the torque about the axis through contact point is zero and as there is no external torque is acting on the body its angular moment this axis will remain conserved. Hence the angular momentum of the body about the axis passing through contact point remains Lo throughout the motion.

When the body starts pure rolling with velocity v (say) the contact point is momentarily at rest and we can say that the body is rotating about this instantaneous axis of rotation with angular velocity  = v/R.

I' = I + mR2

Hence angular moment of the body about instantaneous axis of rotation is

L = I'w = (I + mR2)*(v/R)

Thus according to law of conservation of angular momentum we have

L = L0

Or
Or

Now for sphere thus substituting in above equation we get

Alternative method

Friction force will be mg and thus the torque about the axis from center will be - mgR
Thus angular acceleration about axis through center will be

If the rolling starts in time t than the angular velocity after time t will be

Or ------------------------ (1)

The friction force (forward) is mg

Thus the acceleration of the sphere will be g

Thus velocity of the center of sphere after time t will be

V = 0 + g*t

Now as for pure rolling v = R we get using (1)

Or

Or

Or

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