A high-wire walker always attempts to keep his center of mass over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of $70.0\text{kg}$and a rotational inertia of about the wire.What is the magnitude of his angular acceleration about the wire if his com is $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$to the right of the wire and (a) he carries no pole and (b) the $\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{kg}}\mathbf{}$pole he carries has its com 10 cm to the left of the wire?

As the torque is the product of the force and distance from the pivot to the point of application of force, we can use the formula for torque in terms of angular acceleration and moment of inertia and in terms of distance and force to find the magnitude of angular acceleration. Here, this force is the gravitational force of the walker’s center of mass.

1. Mass of walker$\text{M}=70.0\text{kg}$
2. Rotational inertia of walker$\text{I}=15.0\text{kg}.{\text{m}}^2$
3. Distance$r=5.0\text{cm}=5.0\times {10}^{-2}\text{m}$

$\tau =\text{I}\alpha$

And, in terms of distance and force,

$\tau =\text{rF}$

Therefore,

$\text{I}\alpha =\text{rF}$

But,$\text{F}=\text{mg}$so,

$\text{I}\alpha =\text{rmg}$

$\alpha =\frac{\text{rmg}}{\text{I}}$

$\alpha =\frac{\left(5.0\times {10}^{-2}\text{m}\right)\left(70.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{15{\text{kgm}}^2}$

$\alpha =2.3\text{rad}/{\text{s}}^2$

Therefore, the magnitude of angular acceleration about the wire if his com is $5.0\text{cm}$to the right of the wire and carries no pole is $2.3\text{rad}/{\text{s}}^2$

${\tau }_{net}=\text{I}\alpha$

$\alpha =\frac{{\tau }_{net}}{\text{I}}$

$\alpha =\frac{{\tau }_{walker}-{\tau }_{pole}}{\text{I}}$

$\alpha =\frac{{\text{r}}_{com}\text{mg}-\text{r}\text{'}\text{m}\text{'}\text{g}}{\text{I}}$

$\alpha =\frac{\left(5.0\times {10}^{-2}\text{m}\right)\left(70.0\text{kg}\right)\left(\frac{9.8\text{m}}{{\text{s}}^2}\right)-\left(10\times {10}^{-2}\text{m}\right)\left(14.0\text{kg}\right)\left(\frac{9.8\text{m}}{{\text{s}}^2}\right)}{15{\text{kgm}}^2}$

$\alpha =1.4\frac{\text{rad}}{{\text{s}}^2}$

Therefore, magnitude of angular acceleration about the wire if his com is to the right of the wire and the pole he carries its com is 10 cm to the left of the wire is

$1.4\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.$

Final Statement:

We can find the magnitude of angular acceleration about the wire using the formula for torque in terms of angular acceleration and moment of inertia and in terms of distance and force.