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{excerpt}An extended object in which the distance between any two points in the object is constant in time.{excerpt}

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h2.  Motivation of Concept

Many everyday objects are essentially rigid bodies.  Any object which does not significantly deform in its everyday use can be treated as a rigid body.  Some examples are ceramic cups, keys, wooden chairs, and hockey pucks.  


These objects should be contrasted with other objects that significantly deform when forces are applied.  Examples are handbags, unrolled newspapers, cords, and beanbags.

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h2. Effects of External Forces on Rigid Bodies

h4. Pure Translational Acceleration

Forces applied to an initially non-rotating rigid body in such a way that their [line of action] passes through the body's center of mass will produce pure translation of the rigid body.  The acceleration of every point in the rigid body will be identical and governed by [Newton's 2nd Law|Newton's Second Law] applied to the entire body:

{latex}\begin{large}\[ \sum \vec{F}_{\rm ext} = M_{\rm tot}\vec{a}\] \end{large}{latex}

where _M_~tot~ is the mass of the entire body.

{table}{tr}{td}!davetranslate.png|width=50%!{td}{tr}{table}

h4.  Translational Acceleration of a Rotating Body

If external forces are applied to a body that is initially rotating about its center of mass with their lines of action passing through the body's center of mass, then the acceleration of the center of mass will still be given by the formula:

{latex}\begin{large}\[ \sum \vec{F}_{\rm ext} = M_{\rm tot}\vec{a}_{\rm cm}\] \end{large}{latex}

The rest of the body will move along with the center of mass, but because of the rotation the acceleration of a given point p will differ from that of the cm.  The acceleration of each point can be found by taking the vector sum of the centripetal acceleration needed to maintain the rotation.

h4.  Pure Rotational Acceleration

Even in the case that all external forces sum to zero, an extended rigid body may experience a change in its state of motion.  If the [torques|torque (one-dimensional)] resulting from the applied forces do not sum to zero, the rigid body will experience an [angular acceleration] about its center of mass, changing its rotational state.  Assuming that the body possesses certain [rotational symmetries|angular momentum (one-dimensional)#rotational symmetries], the angular acceleration of every point in the body will be identical and will be governed by the formula:

{latex}\begin{large}\[ \sum \tau = I\alpha \]\end{large}{latex}

where _I_ is the [moment of inertia] of the body about the axis of the resulting rotation.
{table}{tr}{td}!daverotate.png|width=50%!{td}{tr}{table}


h4.  Combined Rotational and Translational Acceleration

If neither the forces nor the torques sum to zero, the motion of the rigid body can be treated as the sum of the resulting translational acceleration and rotational acceleration found by the methods described in the previous sections.  Since the center of mass is the center of the rotation, its linear acceleration is the most easily obtained.  It is given by:

{latex}\begin{large}\[ \sum \vec{F}_{\rm ext} = M_{\rm tot}\vec{a}_{\rm cm}\] \end{large}{latex}

The angular acceleration of the entire body about the center of mass is:

{latex}\begin{large}\[ \sum \tau = I\alpha \]\end{large}{latex}
{table}{tr}{td}!davetransrotate.png|width=50%!{td}{tr}{table}

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h2. Relevant Models

{children:page=Angular Momentum and Torque|all=true}

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h2. Relevant Examples

See any of the [Examples from Rotation].