183_notes:examples:a_rod_rotating_not_around_its_center [Projects & Practices in Physics]

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===== Example: A Rod Rotating Not Around Its Center  =====

A thin rod with mass 140 grams and 60 cm long rotates at an angular speed of 25 radians/s about an axle that is 20cm from one end of the rod (see Figure in Representations). What is its kinetic energy?


=== Facts ===

Thin rod has a mass of 140 grams 

Thin rod is 60 cm long

Thin rod rotates at an angular speed of 25 radians/s

Rotation occurs around an axle that is 20cm from one end of the rod

=== Assumptions and Approximations ===

The rod has to be sufficiently thin so your not worried about contributions to moment of inertia from parts that are far away.

=== Lacking ===

What is the kinetic energy of the rod?

=== Representations ===

{{course_planning:course_notes:mi3e_09-026.jpg?300|}}

$K_{tot} = K_{trans} + K_{rot} = \frac{1}{2}(Mr^{2}_{CM} + I_{CM})\omega^{2}$

$K_{trans} = \frac{1}{2}Mr^{2}_{CM}$

$K_{rot} = \frac{1}{2}I_{CM}\omega^{2}$

$I = (\frac{1}{12})ML^{2}$ for a thin rod

=== Solution ===

There is translational kinetic energy because the center of mass is moving, and there is rotational energy because there is rotation about the center of mass. 

We know that $(r_{CM} = 0.1 m):$ because this the distance to the centre of mass from the point of rotation: (30cm - 20cm).

We are looking for the kinetic energy for a rigid body rotating about a point not the center of the mass.

The total energy has both translational and rotational components

$K_{tot} = K_{trans} + K_{rot}$

Substitute in equations for rotational kinetic energy and translational kinetic energy

$K_{tot} = \frac{1}{2}(Mr^{2}_{CM} + I_{CM})\omega^{2}$

Substitute in $(\frac{1}{12})ML^{2}$ for L as we are dealing with the inertia for a thin rod.

$K_{tot} = \frac{1}{2}(Mr^{2}_{CM}\;+\;\frac{1}{12}ML^{2})\omega^{2}$

Gather the M's out of both equations so that your equation now looks like:

$K_{tot} = \frac{1}{2}M(r^{2}_{CM}\;+\;\frac{1}{12}L^{2})\omega^{2}$

Insert values for the corresponding variables.

$K_{tot} = \frac{1}{2}(.140\;kg)((.1m^2)+\frac{1}{12}(.6m)^2)(25\;radians/s)^2$

Solve for $K_{tot}$

$K_{tot} = 1.75J$