PHYS101: Introduction to Mechanics

Examp[le 10.13 Calculating the Torque in a Kick

The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is $1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}$, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through $57.3^{\circ}$ (1.00 rad)?

Figure 10.22 The muscle in the upper leg gives the lower leg an angular acceleration and imparts rotational kinetic energy to it by exerting a torque about the knee. $F$ is a vector that is perpendicular to $r$. This example examines the situation.

Strategy

The angular acceleration can be found using the rotational analog to Newton's second law, or $\alpha=\text { net } \tau / I$ . The moment of inertia I is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration $\alpha$ is known, the final angular velocity and rotational kinetic energy can be calculated.

Solution to (a)

From the rotational analog to Newton's second law, the angular acceleration $\alpha$ is

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

Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus

$\begin{aligned} \text { net } \tau &=r_{\perp} F \\ &=(0.0220 \mathrm{~m})(2000 \mathrm{~N}) \\ &=44.0 \mathrm{~N} \cdot \mathrm{m} \end{aligned}$

Substituting this value for the torque and the given value for the moment of inertia into the expression for $\alpha$ gives

$\alpha=\frac{44.0 \mathrm{~N} \cdot \mathrm{m}}{1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}}=35.2 \mathrm{rad} / \mathrm{s}^{2}$

Solution to (b)

The final angular velocity can be calculated from the kinematic expression

$\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$

or

$\omega^{2}=2 \alpha \theta$

because the initial angular velocity is zero. The kinetic energy of rotation is

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$

so it is most convenient to use the value of $\omega^{2}$ just found and the given value for the moment of inertia. The kinetic energy is then

$\begin{aligned} \mathrm{KE}_{\mathrm{rot}} &=0.5\left(1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(70.4 \mathrm{rad}^{2} / \mathrm{s}^{2}\right) \\ &=44.0 \mathrm{~J} \end{aligned}$

Discussion

These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick.