Editing Rolling (section)

=== Forces and acceleration ===
Differentiating the relation between linear and angular ''velocity'', <math>v_\text{c.o.m.} = r\omega</math>, with respect to time gives a formula relating linear and angular ''acceleration'' <math>a = r\alpha</math>. Applying [[Newton's second law]]:

<math display="block">a = \frac{F_\text{net}}{m} = r\alpha = \frac{r\tau}{I}.</math>

It follows that to accelerate the object, both a net force and a [[torque]] are required. When external force with no torque acts on the rolling object‐surface system, there will be a tangential force at the point of contact between the surface and rolling object that provides the required torque as long as the motion is pure rolling; this force is usually [[static friction]], for example, between the road and a wheel or between a bowling lane and a bowling ball. When static friction isn't enough, the friction becomes [[dynamic friction]] and slipping happens. The tangential force is opposite in direction to the external force, and therefore partially cancels it. The resulting [[net force]] and acceleration are:

<math display="block">\begin{align}
  F_\text{net} &=
    \frac{F_\text{external}}{1 + \frac{I}{mr^2}} =
    \frac{F_\text{external}}{1 + \left({r_\text{g}}/{r}\right)^2} \\[1ex]
             a &= \frac{F_\text{external}}{m + {I}/{r^2}}
\end{align}</math>

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{{show
|Derivation
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Assume that the object experiences an external force <math>F_\text{external}</math> which exerts no torque (it has 0 [[moment arm]]), static friction at the point of contact (<math>F_{\text{friction}}</math>) provides the torque and the other forces involved cancel. <math>F_{\text{friction}}</math> is tangential to the object and surface at the point of contact and opposite in direction to <math>F_\text{external}</math>. Using the [[sign convention]] by which this force is positive, the net force is:

{{NumBlk|:|<math>F_\text{net} = F_\text{external} - F_\text{friction}</math>|{{EquationRef|1}}}}

Because there is no slip, <math>r\alpha=a</math> holds. Substituting <math>\alpha</math> and <math>a</math> for the linear and rotational version of [[Newton's second law]], then [[equation solving|solving]] for <math>F_\text{friction}</math>:
<math display="block">\begin{align}
                r\frac{\tau}{I} &= \frac{F_\text{net}}{m} \\
  r\frac{rF_\text{friction}}{I} &= \frac{F_\text{net}}{m} \\
              F_\text{friction} &= \frac{IF_\text{net}}{mr^2} \\
\end{align}</math>

Expanding <math>F_\text{friction}</math> in {{EquationNote|1|(1)}}:
<math display="block">\begin{align}
  F_{\text{net}} &= F_\text{external} - \frac{IF_\text{net}}{mr^2} \\
                 &= F_\text{external} - \frac{I}{mr^2}F_\text{net} \\
                 &= \frac{F_\text{external}}{1 + \frac{I}{mr^2}} \\
\end{align}</math>

The last equality is the first formula for <math>F_\text{net}</math>; using it together with Newton's second law, then [[reduction (mathematics)|reducing]], the formula for <math>a</math> is obtained:
<math display="block">\begin{align}
  a &= \frac{F_\text{net}}{m} \\
    &= \frac{\left(\frac{F_\text{external}}{1 + \frac{I}{mr^2}}\right)}{m} \\
    &= \frac{F_\text{external}}{m\left(1 + \frac{I}{mr^2}\right)} \\
    &= \frac{F_\text{external}}{m + {I}/{r^2}} \\
\end{align}</math>

The [[radius of gyration]] can be incorporated in the first formula for <math>F_\text{net}</math> as follows:
<math display="block">\begin{align}
  r_\text{g}   &= \sqrt{\frac{I}{m}} \\
  r_\text{g}^2 &= \frac{I}{m} \\
  \frac{I}{mr^2} &= \frac{\left(\frac{I}{m}\right)}{r^2} \\
                 &= \frac{r_\text{g}^2}{r^2} \\
                 &= \left(\frac{r_\text{g}}{r}\right)^2 \\
\end{align}</math>

Substituting the latest equality above in the first formula for <math>F_\text{net}</math> the second formula for it:
<math display="block">F_\text{net} = \frac{F_{\text{external}}}{1 + \left({r_\text{g}}/{r}\right)^2}</math>
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<math>\tfrac{I}{r^2}</math> has dimension of mass, and it is the mass that would have a rotational inertia <math>I</math> at distance <math>r</math> from an axis of rotation. Therefore, the term <math>\tfrac{I}{r^2}</math> may be thought of as the mass with linear inertia equivalent to the rolling object rotational inertia (around its center of mass). The action of the external force upon an object in simple rotation may be conceptualized as accelerating the sum of the real mass and the virtual mass that represents the rotational inertia, which is <math>m+\tfrac{I}{r^2}</math>. Since the work done by the external force is split between overcoming the translational and rotational inertia, the external force results in a smaller net force by the [[dimensionless]] multiplicative factor <math>1/\left(1+\tfrac{I}{mr^2}\right)</math> where <math>\tfrac{I}{mr^2}</math> represents the ratio of the aforesaid virtual mass to the object actual mass and it is equal to <math display="inline">\left({r_\text{g}}/{r}\right)^2</math> where <math>r_\text{g}</math> is the [[radius of gyration]] corresponding to the object rotational inertia in pure rotation (not the rotational inertia in pure rolling). The square power is due to the fact rotational inertia of a point mass varies proportionally to the square of its distance to the axis.

[[File:Rolling Racers - Moment of inertia.ogv|thumb|right|Four objects in pure rolling racing down a plane with no air drag. From back to front: spherical shell (red), solid sphere (orange), cylindrical ring (green) and solid cylinder (blue).

The time to reach the finishing line is entirely a function of the object mass distribution, slope and gravitational acceleration. See [[:File:Rolling Racers - Moment of inertia.ogv|details]], [[:File:Rolling Racers - Moment of inertia.gif|animated GIF version]].]]
In the specific case of an object rolling in an [[inclined plane]] which experiences only static friction, normal force and its own weight, ([[air drag]] is absent) the acceleration in the direction of rolling down the slope is:

<math display="block">a = \frac{g\sin(\theta)}{1+\left({r_\text{g}}/{r}\right)^2}</math>

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{{show
|Derivation
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Assuming that the object is placed so as to roll downward in the direction of the inclined plane (and not partially sideways), the weight can be decomposed in a component in the direction of rolling and a component perpendicular to the inclined plane. Only the first force component makes the object roll, the second is balanced by the contact force, but it does not form an action‐reaction pair with it (just as an object in rest on a table). Therefore, for this analysis, only the first component is considered, thus:

<math display="block">F_\text{external} = gm \sin(\theta)</math>

<math display="block">\begin{align}
a&=\frac{F_\text{net}}{m} \\
&=\frac{\left(\frac{F_{\text{external}}}{1+\left({r_\text{g}}/{r}\right)^2}\right)}{m} \\
&=\frac{\left(\frac{gm\sin(\theta)}{1+\left({r_\text{g}}/{r}\right)^2}\right)}{m} \\
&=\frac{gm\sin(\theta)}{m\left(1+\left({r_\text{g}}/{r}\right)^2\right)} \\
&=\frac{g\sin(\theta)}{1+\left({r_\text{g}}/{r}\right)^2} \\
\end{align}</math>

In the last equality the denominator is the same as in the formula for force, but the factor <math>m</math> disappears because its instance in the force of gravity cancels with its instance due to Newton's third law.
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<math>{r_\text{g}}/{r}</math> is specific to the object shape and mass distribution, it does not depend on scale or density. However, it will vary if the object is made to roll with different radiuses; for instance, it varies between a train wheel set rolling normally (by its tire), and by its axle. It follows that given a reference rolling object, another object bigger or with different density will roll with the same acceleration. This behavior is the same as that of an object in free fall or an object sliding without friction (instead of rolling) down an inclined plane.