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Section 10.6 Kinetic Energy of a Rigid Body

Kinetic energy of a particle of mass moving with speed \(v\) is given by

\begin{equation*} K = \frac{1}{2}mv^2. \end{equation*}

Now, if we have a collection of particles \(m_1\text{,}\) \(m_2\text{,}\) \(\cdots\text{,}\) with speeds \(v_1\text{,}\) \(v_2\text{,}\) \(\cdots\text{,}\) the total kinetic energy will be just the sum.

\begin{equation} K_\text{tot} = \frac{1}{2}\sum_{i}m_i v_i^2.\label{eq-total-kinetic-energy}\tag{10.6.1} \end{equation}

Similar to how the total angular momentum separated into angular momentum of CM and angular momentum about the CM, similar calculations lead to the separation of total kinetic energy can be separated into a translational motion of the CM and a rotation about an axis through the CM. This rotation energy is called rotational kinetic energy.

\begin{equation} K_\text{tot} = K_\text{cm} + K_\text{rot}\label{eq-total-kinetic-energy-as-sum-cm-plus-rot}\tag{10.6.2} \end{equation}

In the following section we will show that \(K_\text{rot}\) is given by

\begin{equation*} K_\text{rot} = \frac{1}{2}\,\vec\omega \cdot \vec L. \end{equation*}

This simplifies when we work with principal axes. Then, we can write this as

\begin{equation} K_\text{rot} = \frac{1}{2} I_{xx}\omega_x^2 + \frac{1}{2} I_{yy}\omega_y^2 + \frac{1}{2} I_{zz}\omega_z^2.\tag{10.6.3} \end{equation}

Using L_i = I_{ii}\omega_i applicable for principal axes, we can write it using angular momentum as well.

\begin{equation*} K_\text{rot} = \frac{1}{2} \frac{L_x^2}{I_{xx}} + \frac{1}{2} \frac{L_y^2}{I_{yy}} + \frac{1}{2} \frac{L_z^2}{I_{zz}}. \end{equation*}

Subsection 10.6.1 Kinetic Energy About Center of Mass

We found in our last chapter that we can separate translational motion of the center of mass motion and motion about the center of mass by replacing velocity \(\vec v_i\) by sum of velocity \(\vec v'_i\) with respect to CM and velocity \(V_\text{cm}\) of the CM.

\begin{equation*} \vec v_i = \vec V_\text{cm} + \vec v'_i. \end{equation*}

The calculation of separation gives

\begin{align*} K_\text{tot} \amp = \frac{1}{2}\sum_{i}m_i v_i^2 = \frac{1}{2}\sum_{i}m_i \vec v_i \cdot \vec v_i \\ \amp = \frac{1}{2}\sum_{i}m_i (\vec V_\text{cm} + \vec v'_i) \cdot (\vec V_\text{cm} + \vec v'_i) \\ \amp = \frac{1}{2}\vec V_\text{cm} \cdot \vec V_\text{cm} \sum_{i}m_i + \vec V_\text{cm}\cdot \sum_{i}m_i \vec v'_i + \frac{1}{2}\sum_{i}m_i\vec v'_i\cdot \vec v'_i \\ \amp = \frac{1}{2} M V_\text{cm}^2 + \frac{1}{2}\sum_{i}m_i\vec v'_i\cdot \vec v'_i. \end{align*}

In the last step I used the fact that \(\sum_{i}m_i \vec v'_i=0\) since \(\vec v'_i\) are velocities with respect to the CM and have replaced total mass by \(M\text{.}\) The first part is kinetic energy of translation of the CM and the second part is the rotational kinetic energy.

\begin{align} \amp K_\text{cm} = \frac{1}{2} M V_\text{cm}^2. \tag{10.6.4}\\ \amp K_\text{rot} = \frac{1}{2}\sum_{i}m_i\vec v'_i\cdot \vec v'_i. \tag{10.6.5} \end{align}

By making use of the relation, \(\vec v' = \vec\omega \times \vec r'\text{,}\) for each particle, we can write \(K_\text{rot}\) in terms of angular velocity and angular moementum. Thus,

\begin{align*} K_\text{rot} \amp = \frac{1}{2}\sum_{i}m_i\vec v'_i\cdot \vec v'_i\\ \amp = \frac{1}{2}\sum_{i}m_i(\vec\omega \times \vec r'_i)\cdot (\vec\omega \times \vec r'_i)\\ \amp = \frac{1}{2}\sum_{i}m_i\vec\omega \cdot [ \vec r'_i \times (\vec\omega \times \vec r'_i)]\\ \amp = \frac{1}{2}\vec\omega \cdot \sum_{i}m_i [ \vec r'_i \times (\vec\omega \times \vec r'_i)] . \end{align*}

The expression \(\sum_{i}m_i [ \vec r'_i \times (\vec\omega \times \vec r'_i)]\) is angular momentum \(\vec L\) about CM by (10.2.3). Therefore,

\begin{equation} K_\text{rot} = \frac{1}{2}\,\vec\omega \cdot \vec L.\tag{10.6.6} \end{equation}

We can replace \(\vec L\) by \({\bf I}\vec \omega\text{,}\) where \({\bf I}\) is full moment of inertia, giving us

\begin{equation} K_\text{rot} = \frac{1}{2}\,\vec\omega \cdot ( {\bf I}\vec \omega ).\tag{10.6.7} \end{equation}

We know that \({\bf I}\) simplifies when we use principal axes of the rotating body. Therefore, in a coordinate system along principal axes we can write kinetic energy in terms of components of angular velocity as

\begin{equation} K_\text{rot} = \frac{1}{2} I_{xx}\omega_x^2 + \frac{1}{2} I_{yy}\omega_y^2 + \frac{1}{2} I_{zz}\omega_z^2.\tag{10.6.8} \end{equation}

Often, it is useful to think of this relation in terms of angular momentum components rather than angular velocity components. Since in principal axes, \(L_{i} = I_{ii}\omega_i\) for \(i=x,y,z\text{,}\) rotational kinetic energy can also be written as

\begin{equation} K_\text{rot} = \frac{1}{2} \frac{L_x^2}{I_{xx}} + \frac{1}{2} \frac{L_y^2}{I_{yy}} + \frac{1}{2} \frac{L_z^2}{I_{zz}}. \tag{10.6.9} \end{equation}

From this expression of \(K_\text{rot}\text{,}\) we can see that the principal axis about which moment of inertia is largest would tend to minimize kinetic energy. This leads to more stability for rotations around principal axis of largest \(I\text{.}\) If you start a rotation around an axis that has smallest moment of inertia, then, over time, the body will wobble and tend to settle in rotation about axis of largest moment of inertia.

That is why rotation of a thin disk about axis perpendicular to the disk is more stable than one about an axis in the plane of the disk.

Figure 10.6.1.

A rod of mass \(M\) and length \(l\) is poivoted at one end as shown in Figure 10.6.3. At a particular instant angular speed is \(\omega\text{.}\) Find kinetic energy of the rod in two ways: (a) rotation about \(P\text{,}\) (b) sum of translation of CM and rotation about the CM.

Figure 10.6.3.

Follow tge suggestio in the problem statement.


\(\frac{1}{6}Ml^2 \omega^2\text{.}\)


(a) For rotation about \(P\text{,}\) we need moment of inertia about \(P\text{,}\) which is \(I=\frac{1}{3}Ml^2\text{.}\) Therefore, kinetic energy will be

\begin{equation*} K = \frac{1}{2}I \omega^2 = \frac{1}{6}Ml^2 \omega^2. \end{equation*}

(b) Now, we have \(I_0 = \frac{1}{12}Ml^2\text{.}\) We also need the speed of CM. Since CM is moving in a circle of radius \(\frac{l}{2}\) with angular speed \(\omega\text{,}\) we get

\begin{equation*} V_\text{cm} = \frac{l}{2}\,\omega. \end{equation*}

Therefore, kinetic energy will be sum of CM translational energy and rotation about the CM.

\begin{align*} K \amp = \frac{1}{2}M V_\text{cm}^2 + \frac{1}{2}I_0 \omega^2 \\ \amp = \frac{1}{2}M \frac{l^2}{4}\,\omega^2 + \frac{1}{2} \frac{1}{12}Ml^2 \omega^2 \\ \amp = Ml^2 \omega^2\left( \frac{1}{8} + \frac{1}{24} \right) = \frac{1}{6}Ml^2 \omega^2. \end{align*}

A cylinder of mass \(M\) and radius \(R\) is rolling without slipping with center moving at speed \(V\text{.}\) What is its kinetic energy?

Figure 10.6.5.

Its not just \(\frac{1}{2}MV^2\text{.}\)




Kinetic energy will be sum of translational kinetic energy of center of mass and rotation about an axis through the center of mass. To find the rotation velocity, we attach a moving coordinate system in the body with origin at the center of mass as shown in Figure 10.6.6.

Figure 10.6.6.

From the figure, it is easy to tell that angular speed \(\omega\) will be related to speed \(V\) with which the cylider rolls on the surface.

\begin{equation*} \omega = \frac{\Delta \theta}{\Delta t} = \frac{1}{R}V. \end{equation*}

Hence, kinetic energy will be

\begin{align*} K \amp = \frac{1}{2}MV^2 + \frac{1}{2} I_0 \omega^2 \\ \amp = \frac{1}{2}MV^2 + \frac{1}{2} \frac{1}{2}MR^2\omega^2 = \frac{3}{4}MV^2. \end{align*}

A cylinder of mass \(M\) and radius \(a\) is rolling without slipping with center moving at speed \(V\) on a cylindrical surface of radius of curvature \(R\) as shown in Figure 10.6.8. (a) What is its kinetic energy? (b) What is angular momentum about center of the center of the cylindrical surface?

Figure 10.6.8.

For angular velocity, think of rotation about instantaneously at-rest line as axis of rotation.


(a) \(\frac{3}{4}MV^2\text{,}\) (b) \(MV\left( R - \frac{a}{2}\right)\) coming out of page.

Solution 1 (a)

(a) Kinetic energy with respect to the fixed axis will be sum of translational kinetic energy of center of mass and rotation about an axis through the center of mass.

To find angular velocity \(\omega\) in terms of speed of CM \(V\text{,}\) we use instantaneous axis of rotation passing through point P in Figure 10.6.9.

Figure 10.6.9.

With velocity \(V\) of the CM about instantaneously at-rest point P at a distance \(a\) from P, angular velocity will be

\begin{equation*} \omega = \frac{V}{a}. \end{equation*}

Therefore, kinetic energy of the cylinder will be

\begin{align*} K \amp = \frac{1}{2}MV^2 + \frac{1}{2} I_0 \omega^2 \\ \amp = \frac{1}{2}MV^2 + \frac{1}{2} \frac{1}{2}Ma^2\omega^2 = \frac{3}{4}MV^2. \end{align*}
Solution 2 (b)

Both the angular momentum associated with the motion of the CM and the angular momentum about an axis through CM will be coming out of the page in Figure 10.6.9. Let that axis be \(z\)-axis of body-fixed coordinate system and \(z'\) of the fixed coordinate system. Then, angular momentum about \(O'\) will be

\begin{equation*} \vec L = MV(R-a)\hat k + I_{zz}\omega\hat k. \end{equation*}

With \(I_{zz} = \frac{1}{2}Ma^2\) and \(\omega=V/a\text{,}\) this simlifies to

\begin{equation*} L_z = MV(R-a) + \frac{1}{2}MaV = MV\left( R - \frac{a}{2}\right). \end{equation*}

A cylinder of mass \(M\text{,}\) radius \(R\) and height \(H\) has a tiny hole through the center. When a pin is inserted in the hole, cylinder can rotate about the pin freely as shown in Figure 10.6.11.

Holding the pin at one end, the cylinder is rolled on a flat surface such that cylinder rolls without sliding with its center moving at speed \(V\text{.}\) Find kinetic energy of the cylinder.

Figure 10.6.11.

No hint.




For the sake of simplification, let us pick a moment shown in Figure 10.6.12 where the principal axis \(Oz\) is in the direction of fixed axis \(O'x'\text{.}\)

Figure 10.6.12.

By looking at motion of CM at \(x'=\frac{L}{2},\ y'=0,\ z'=R\) about instantaneous axis or rotation \(O'A\text{,}\) we see that angular velocity pointed towards negative \(z\)-axis with magnitude

\begin{equation*} \omega = V/R. \end{equation*}

Therefore, we have kinetic energy

\begin{align*} K \amp = \frac{1}{2}MV^2 + \frac{1}{2} I_{zz} \omega^2 \\ \amp = \frac{1}{2}MV^2 + \frac{1}{2} \frac{1}{2}MR^2 (V/R)^2 = \frac{3}{4}MV^2. \end{align*}

A cone of mass \(M\text{,}\) height \(h\) and apex angle \(2\alpha\) is placed on a flat surface and rolled wwithout slipping around its apex. The line touching a flat plane sweeps out \(\Omega\) rad/sec about the vertical axis as shown in Figure 10.6.14. Find kinetic energy of the cone.

Figure 10.6.14.

Center of mass of a cone is at \(\frac{3}{4}h\) from the apex and formulae for principal moments of inertia are \(I_1=I_2=\frac{3}{20}M\left(R^2 + \frac{1}{4}h^2 \right)\text{,}\) \(I_3=\frac{3}{10}MR^2\text{.}\)


Find the radius of the circle in which CM moves around \(O'A\) to get angular speed.


\(\frac{3}{40}\,Mh^2\Omega^2\left( 1 + 5 \cos^2 \alpha \right)\text{.}\)


Figure 10.6.15 shows fixed and moving coordinates. At the instant depicted in the figure, instantaneous axis of rotation is in the direction of \(O'A\) in \(O'x'y'\) plane. The origin of the moving coordinate system is at the center of mass with \(\overline{O'O}=\frac{3}{4}h\text{.}\)

Figure 10.6.15.

The CM will go around axis \(O'A\) in a circle of radius \(b\) given by

\begin{equation*} b = a\,\sin\alpha. \end{equation*}

Angular speed of rotation about \(O'A\) will be

\begin{equation*} \omega = \frac{V}{b} = \frac{V}{a\,\sin\alpha}. \end{equation*}

Figure 10.6.16.

Now, we work out components of angular velocity vector with respect to principal axes attached to moving cone so that we can use diagonal \({\bf I}\text{.}\)

Since direction of angular velocity vector is in \(O'A\) direction, the components of \(\vec \omega\) in the moving coordinate system will be

\begin{equation*} \omega_x = -\omega\,\sin\alpha , \omega_y = 0, \omega_z = \omega\,\cos\alpha. \end{equation*}

Figure 10.6.17.

Therefore, kinetic energy about \(O'\) will be

\begin{align} K \amp = \frac{1}{2} MV^2 + \frac{1}{2} I_{xx}\omega_x^2 + \frac{1}{2} I_{zz}\omega_z^2\notag\\ \amp = \frac{1}{2} MV^2 + \frac{1}{2}\times \frac{3}{20}M\left(R^2 + \frac{1}{4}h^2 \right) \omega^2\sin^2\alpha\notag\\ \amp \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \frac{1}{2}\times \frac{3}{10}MR^2 \omega^2\,\cos^2\alpha \label{eq-kinetic-energy-cone-prelim}\tag{10.6.10} \end{align}

We can express kinetic energy in \(\Omega\) by working out \(V\) of the CM in terms of \(\Omega\text{.}\) As showin in Figure 10.6.18, CM moves in a circle of radius \(a\cos\alpha\text{.}\) Therefore,

\begin{equation*} V = \Omega a\cos\alpha. \end{equation*}

Figure 10.6.18.

Using \(V\) in \(\omega\) we get

\begin{equation*} \omega = \frac{V}{a\sin\alpha} = \frac{\Omega a\cos\alpha}{a\sin\alpha} = \Omega\,\cot\alpha. \end{equation*}

Use these along with \(R = h\tan\alpha\) and \(a = 3h/4\) to convert Eq. (10.6.10) to

\begin{equation*} K = \frac{3}{40}\,Mh^2\Omega^2\left( 1 + 5 \cos^2 \alpha \right). \end{equation*}