Question

84. A thin rod of mass$\mathbf{0}\mathbf{.}\mathbf{630}\mathbf{}\mathit{k}\mathit{g}$ and length$\mathbf{1}\mathbf{.}\mathbf{24}\mathbf{}\mathit{m}$ is at rest, hanging vertically from a strong, fixed hinge at its top end. Suddenly, a horizontal impulsive force $\mathbf{14}\mathbf{.}\mathbf{7}\widehat{\mathbf{i}}\mathit{N}$is applied to it. (a) Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass and (b) the horizontal force the hinge exerts. (c) Suppose the force acts at the midpoint of the rod. Find the acceleration of this point and (d) the horizontal hinge reaction force. (e) Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion.

Short answer

$0.827m$The solution is

a)$a_{CM}=35.0m/s^2$

b)$H_X=7.35\widehat{i}N$

c)$a_{CM}=17.5m/s^2$

d)$H_x=-3.68\widehat{N}$

e) The center of percussion (the point where the hinge exerts no force is

Step-by-step solution

of 10: Conceive

The moment of inertia of the rod is used to conceptualize the situation.

of 10: CATEGORIZE

This problem is classified as a rigid body torque equation problem.

of 10: Given Information

Given that

Mass of the rod$m=0.630kg$

Length of the rod$L=1.24m$

Magnitude of force$F=14.7N$

of 10: Free body diagram

Consider the free body diagram as illustrated for the given problem.

of 10: Balancing the torque

The sum of torques about the chosen pivot is, as shown in the diagram.

$\begin{array}{l}\Sigma \tau =I\alpha \\ \Sigma \tau =I\left(\frac{a}{R}\right)\end{array}$

$\begin{array}{l}Fl=\left(\frac{1}{3}m{L}^2\right)\left(\frac{a_{CM}}{\frac{L}{2}}\right)\\ =\left(\frac{2}{3}mL\right)a_{CM}\cdots \cdots \left(1\right)\end{array}$

of 10: Acceleration of centre of mass

$\begin{array}{l}a)l=L\\ =1.24m\end{array}$

The equation (1) becomes in this situation.

$\begin{array}{c}{a}_{CM}=\frac{3F}{2m}\\ =\frac{3(14.7N)}{2(0.630kg)}\\ =35.0m/s^2\end{array}$

of 10: Horizontal force exerted by the hinge

$\begin{array}{l}\left(b\right)\Sigma F_x=m{a}_{CM}\\ F+H{a}_x=m{a}_{CM}\\ H_x=m{a}_{\text{CM}}-F\end{array}$

$\begin{array}{l}=(0.630kg)(35.0m/s^2)-14.7N\\ =7.35N\\ H_x=7.35\widehat{i}N\end{array}$

of 10: Acceleration of the midpoint of the rod

$\begin{array}{l}c)l=\frac{1}{2}\\ =0.620m\end{array}$

The equation (1) becomes in this situation.

$\begin{array}{c}{a}_{CM}=\frac{3F}{4m}\\ =\frac{3(14.7N)}{4(0.630kg)}\\ =17.5m/s^2\end{array}$

of 10: Horizontal hinge reaction force

$\begin{array}{l}d)\Sigma F_x=m{a}_{CM}\\ H_x=m{a}_{CM}-F\\ H_x=(0.630kg)(17.5m/s^2)-14.7N\\ =-3.68N\end{array}$

of 10: Point where impulse will applied

(e) If$H_x=0$, then

$\begin{array}{l}\Sigma F_x=m{a}_{CM}\\ F=m{a}_{CM}\\ a_{CM}=\frac{F}{m}\end{array}$

The equation (1) becomes in this situation.

$\begin{array}{l}Fl=\left(\frac{2}{3}mL\right)\left(\frac{F}{m}\right)\\ l=\left(\frac{2}{3}\right)(1.24m)\\ =0.827m\end{array}$

FINAL ANALYSIS: From the top, the center of percussion (the point where the hinge exerts no force) is$0.827m$