The forearm in Fig. 8–46 accelerates a 3.6-kg ball at \({\bf{7}}{\bf{.0}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) by means of the triceps muscle, as shown. Calculate (a) the torque needed and (b) the force that must be exerted by the triceps muscle. Ignore the mass of the arm.

FIGURE 8-46

Problems 35 and 36

Torque is the product of the moment of inertia and the angular acceleration. For this problem, you have to find the relationship between linear acceleration and angular acceleration.

The mass of the ball is \(m = 3.6\;{\rm{kg}}\).

The length of the forearm is \(r = 31\;{\rm{cm}} = 0.31\;{\rm{m}}\).

The translational acceleration of the ball is \(a = 7.0\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

The perpendicular distance of the triceps from the axis of rotation is \({r_1} = 2.5\;{\rm{cm}} = 0.025\;{\rm{m}}\).

Let \(\tau \) torque be needed to produce such acceleration, and the force exerted by the triceps muscle is \(F\).

You can consider the joint as a point mass.

Part (a)

The moment of inertia of the ball relative to the axis of rotation is \(I = m{r^2}\).

Now, the angular acceleration of the ball is \(\alpha = \frac{a}{r}\).

Therefore, the torque on the ball is

\(\begin{align}\tau &= I\alpha \\ &= m{r^2} \times \frac{a}{r}\\ &= mra\end{align}\).

Now, substituting the values in the above equation, you get

\(\begin{align}\tau &= mra\\ &= \left( {3.6\;{\rm{kg}}} \right) \times \left( {0.31\;{\rm{m}}} \right) \times \left( {7.0\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ &= 7.8\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, the required value of the torque is \(7.8\;{\rm{N}} \cdot {\rm{m}}\).

Part (b)

Now, to produce that amount of torque,

\(\begin{align}\tau &= F{r_1}\\7.8\;{\rm{N}} \cdot {\rm{m}} &= F \times 0.025\;{\rm{m}}\\F &= 312\;{\rm{N}}\\ \approx 310\;{\rm{N}}\end{align}\).

Hence, the force exerted by the triceps muscle is \(310\;{\rm{N}}\).