# Rigid body/moment of inertia problem

I have a homework assignment about rigid body dynamics. Take a disc of radius $r=2m$ with uniform mass density $\rho=1$ $kg/m^2$ in the x-y plane, resting in an inertial frame. At some instant, a force of $F = (0,0,1)kN$ is applied at the point $A=(0,r,0)$. What's the acceleration of the point $A$ at that instant?

This should be straightforward computation, but this is new to me and I think I'm making a fundamental mistake somewhere. Could you help me find it? In particular I'm not sure I'm treating the angular velocity $\omega$ correctly. I can assume the equations of motion for a rigid body, as well as the equation for the acceleration of an arbitrary point of the body given the acceleration of the center of mass and the angular acceleration.

The disc is just a circle of homogenous mass $\rho = 1$, so

$$M = \rho \int_{R=0}^r\int_{\theta=0}^{2\pi} R d\theta dR = \left.\frac{\rho 2\pi R^2}{r}\right|_{R=0}^2 = \left.\pi R^2\right|_{R=0}^r = 4\pi~\mbox{kg}$$

where though the the problem gives $r=2$ I'm just going to leave it as symbols. Without integration, I suppose we might have noticed that $M = \rho \pi r^2$. Now we compute the tensor of inertia,

$$\begin{pmatrix} \int_V \rho x_2^2 +x_3^2 dv & -\int_V \rho x_1 x_2 dv& -\int_V \rho x_1x_3dv \\ -\int_V \rho x_1 x_2dv & \int_V \rho x_1^2 +x_3^2dv & -\int_V \rho x_2 x_3dv \\ -\int_V \rho x_1 x_3 dv& -\int_V \rho x_2 x_3dv & \int_V \rho x_1^2 +x_2^2dv \end{pmatrix}$$

Since our object is symmetric about the x-y axes, we can eliminate the cross-moments. Since our object has no mass in the z direction, we can eliminate $x_3$ in the tensor, so

\begin{align*} I &= \begin{pmatrix} \int_V \rho x_2^2 dv & 0& 0 \\ 0 & \int_V \rho x_1^2 dv & 0 \\ 0 & 0 & \int_V \rho x_1^2 +x_2^2dv \end{pmatrix} \\ &= \begin{pmatrix} \rho \int r^3 sin^2\theta drd\theta & 0& 0 \\ 0 & \rho \int r^3 cos^2\theta drd\theta & 0 \\ 0 & 0 & \rho \int r^3 dr d\theta \end{pmatrix} \\ &= \begin{pmatrix} \frac{\rho}{2} \int r^3 (1-\cos(2\theta)) drd\theta & 0& 0 \\ 0 & \rho \int r^3 (1+\cos(2\theta)) drd\theta & 0 \\ 0 & 0 & \frac{\pi \rho r^4}{2} \end{pmatrix} \\ &= \begin{pmatrix} \frac{\rho}{2}\left[ \frac{2\pi r^4}{4} - \frac{1}{2} \sin(4\pi)\right] & 0& 0 \\ 0 & \frac{\rho}{2}\left[ \frac{2\pi r^4}{4} + \frac{1}{2} \sin(4\pi)\right] & 0 \\ 0 & 0 & \frac{\pi \rho r^4}{2} \end{pmatrix} \\ &= \begin{pmatrix} \frac{\rho \pi r^4}{4} & 0& 0 \\ 0 & \frac{\rho \pi r^4}{4} & 0 \\ 0 & 0 & \frac{\pi \rho r^4}{2} \end{pmatrix} \end{align*}

The angular momentum vector should be $\begin{pmatrix} \omega_x,& 0,& 0 \end{pmatrix}^T$, taking the cross product of the force vector and the vector from the center of mass to $A$. Now we plug this in to the equations of motion to solve for the linear and angular acceleration, $a_c$ and $\alpha$. They are,

\begin{align*} (Mv_c)' &= S \implies & a_c = \frac{1}{M}\begin{pmatrix} 0,& 0,& F \end{pmatrix}^T \\ (I\omega)' &= M_c = \tilde r \times F \implies & \dot \omega = \frac{r}{I_{11}} e_y\times |F|e_z = \frac{r|F|}{I_{11}} e_x \end{align*}

We should now be able to plug this into the equation for angular acceleration (where the angular velocity at the instant of impulse is $0$ so we ignore it),

\begin{align*} a_A &= a_c + \alpha \times e_y \\ &= \frac{|F|}{M}e_z + \dot \omega \times(r e_y) \\ &= |F|(\frac{1}{M} + \frac{r^2}{I_{11}} )e_z \\ &= |F|(\frac{1}{\rho \pi r^2} + \frac{4}{\rho \pi r^2} )e_z \end{align*}

Which I guess seems natural enough. But by newton's formula we would have

\begin{equation*} M a_A = 5 |F| \end{equation*}

But does it make sense that the force should be $5F$? Not really...

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I'm afraid your result is just right, although you are making things overly complicated by considering the full three dimensional rigid body dynamics... The center of mass will start accelerating in the direction of the force with $a=F/M$, plus the whole disk will start turning around a diameter, with an angular acceleration of $\alpha = F r/I$. You have already calculated $I= \rho \pi r^4/4 = M r^2/4$. So the acceleration of your point will be $a+r\alpha = \frac{F}{M}(1+4)=5\frac{F}{M}$.
It may help understand what's going on if you think of what's happening to the point at the other side of the disc, $(0,-r,0)$ in your case, you'll notice that it is accelerating in the opposite direction, $a-r\alpha = \frac{F}{M}(1-4)=-3\frac{F}{M}$. While the disc will also start moving in the direction of the force as a whole, it is also rotating around its center of mass, and the acceleration due to rotation is four times larger than that due to translation. So it is the rotation that is making the acceleration of your point five times larger than you expected.