Question

Assume that a 1.00-kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, as shown in Fig. 8–46. The ball is accelerated uniformly from rest to 8.5 m/s in 0.38 s, at which point it is released. Calculate (a) the angular acceleration of the arm and (b) the force required for the triceps muscle. Assume that the forearm has a mass of 3.7 kg, and it rotates like a uniform rod about an axis at its end.

FIGURE 8-46

Problems 35 and 36

Short answer

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

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

Step-by-step solution

Concept

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.

Given data

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.

Calculation of part (a)

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 \(72.16\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\).

Calculation of part (b)

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 \(619.49\;{\rm{N}}\).