91Ó°ÊÓ

The torque is the product of the moment of inertia and the angular acceleration. For this problem, the total torque on the forearm and the ball is equal to the torque by the force of the triceps muscle.

The mass of the ball is \(m = 1.0\;{\rm{kg}}\).

The mass of the forearm is \(M = 3.7\;{\rm{kg}}\).

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

The initial linear velocity of the ball is \({v_1} = 0\).

The final linear velocity of the ball is \({v_2} = 8.5\;{\rm{m/s}}\).

The ball takes time \(\Delta t = 0.38\;{\rm{s}}\) to gain the speed \({v_2} = 8.5\;{\rm{m/s}}\).

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

Let \(\alpha \) be the angular acceleration of the forearm, and the force exerted by the triceps muscle is \(F\).

You can consider the tricep as a point mass and the forearm as a uniform rod.

Part (a)

The linear acceleration of the ball is \(a = \frac{{{v_2} - v_1^{}}}{{\Delta t}}\).

Now, the angular acceleration is

\(\begin{align}\alpha &= \frac{a}{r}\\ &= \frac{1}{r}\frac{{{v_2} - v_1^{}}}{{\Delta t}}\\ &= \frac{1}{{0.31\;{\rm{m}}}}\frac{{8.5\;{\rm{m/s}} - 0}}{{0.38\;{\rm{s}}}}\\ &= 72.16\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\end{align}\).

Hence, the angular acceleration of the arm is \(72.16\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\).

The moment of inertia of the ball relative to the axis of rotation is \({I_1} = m{r^2}\).

The moment of inertia of the forearm relative to the axis of rotation is \({I_2} = \frac{1}{3}M{r^2}\).

Now, the torque on the arm-ball system is:

\(\begin{align}\tau &= \left( {{I_1} + {I_2}} \right)\alpha \\ &= \left( {m{r^2} + \frac{1}{3}M{r^2}} \right)\alpha \\ &= \left( {m + \frac{M}{3}M} \right){r^2}\alpha \end{align}\).

Now, to produce that amount of torque,

\(\begin{align}\tau &= F{r_1}\\F{r_1}{\rm} &= \left( {m + \frac{M}{3}M} \right){r^2}\alpha \\F \times \left( {0.025\;{\rm{m}}} \right) &= \left( {1.0\;{\rm{kg}} + \frac{{3.7\;{\rm{kg}}}}{3}} \right) \times {\left( {0.31\;{\rm{m}}} \right)^2} \times \left( {72.16\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right)\\F &= 619.49\;{\rm{N}}\end{align}\).

Hence, the required force by the triceps muscles is \(619.49\;{\rm{N}}\).