In a judo foot-sweep move, you sweep your opponent’s left foot out from under him while pulling on his gi (uniform) toward that side. As a result, your opponent rotates around his right foot and onto the mat. Figure $\mathbf{10}\mathbf{-}\mathbf{44}\mathbf{}$shows a simplified diagram of your opponent as you face him, with his left foot swept out. The rotational axis is through point $\mathit{O}$. The gravitational force ${\overrightarrow{\mathbf{F}}}_{\mathbf{g}}$ on him effectively acts at his center of mass, which is a horizontal distance$\mathbf{d}\mathbf{=}\mathbf{}\mathbf{28}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}$ from point $\mathbf{\text{O}}$. His mass is $\mathbf{\text{70Kg}}$, and his rotational inertia about point $\mathbf{\text{O}}$ is $\mathbf{65}{\mathbf{\text{\hspace{0.17em}kg.m}}}^{\mathbf{\text{2}}}$.What is the magnitude of his initial angular acceleration about point $\mathit{O}$ if your pull ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}\mathbf{}$ on his gi is (a) negligible and (b) horizontal with a magnitude of $\mathbf{300}\mathit{N}$and applied at height $\mathbf{h}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.4}\mathbf{\text{\hspace{0.17em}m}}$ ?

1. The mass of opponent is,$\mathrm{m}=70\text{\hspace{0.17em}kg}$.
2. The gravitational acceleration is,$\mathrm{g}=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$
3. ${\mathrm{h}}_1=0.28\text{\hspace{0.17em}m}$
4. The moment of inertia is,$I=65{\text{\hspace{0.17em}kgm}}^{\text{2}}$
5. ${\mathrm{h}}_2=1.4\text{\hspace{0.17em}m}$
6. $F_2=300\text{N}$
   

Let the rotational axis be at point O. The gravitational force will act along the player’s center of mass at some distance. In this situation you can find his initial angular acceleration along point O. But this may happen in two cases. One with the force applied by you is negligible along point O. And second with the force applied by you in horizontal direction along point O i.e., total net torque to be calculated. This is how angular acceleration for both cases can be found. The formulas used are given below.

$\begin{array}{c}{\mathbf{\tau }}_{\mathbf{net}}\mathbf{=}\mathbf{I\alpha }\\ \mathbf{Force}\mathbf{=}\mathbf{mgh}\end{array}$

Here force can be considered as torque,

$\mathrm{Force}=\mathrm{mgh}$

Therefore torque,

$\begin{array}{c}{\mathrm{\tau }}_{\mathrm{net}}=\mathrm{I\alpha }\\ \mathrm{Force}=\mathrm{I\alpha }\end{array}$

Also, moment of inertia is given i.e., $\mathrm{I}=65{\text{\hspace{0.17em}kg.m}}^2$

Therefore,

$\begin{array}{c}\mathrm{\alpha }=\frac{\mathrm{Force}}{\mathrm{I}}\\ \mathrm{\alpha }=\frac{\mathrm{mgh}}{\mathrm{I}}\end{array}$

$\begin{array}{c}\mathrm{\alpha }=\frac{(70\text{\hspace{0.17em}kg})(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\left(0.28\text{\hspace{0.17em}m}\right)}{\left(65{\text{\hspace{0.17em}kg.m}}^{\text{2}}\right)}\\ =2.95\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\\ \approx 3\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Hence the initial angular acceleration about point O when pull is negligible is, $3\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.

Now considering the second case where the force is already mentioned i.e.,$\mathrm{F}=300\text{N}$. We may add both forces to get the angular acceleration along horizontal O point.

Therefore, net torque will be,

$\begin{array}{c}{\mathrm{\tau }}_{\mathrm{net}}=\mathrm{I\alpha }\\ \mathrm{\alpha }=\frac{\mathrm{Force}}{\mathrm{I}}\\ \mathrm{\alpha }=\frac{(70\text{\hspace{0.17em}kg})(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\left(0.28\text{\hspace{0.17em}m}\right)+1.4\text{\hspace{0.17em}m}\times 300\text{\hspace{0.17em}N}}{\left(65{\text{\hspace{0.17em}kg.m}}^{\text{2}}\right)}\\ =9.4\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Hence the initial angular acceleration about point O when pull is horizontal then, $\mathrm{\alpha }=9.4\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.