Example 4
Completion requirements
| Example 4 |
A grandfather clock’s pendulum has a circular weight with a radius of \(5\) cm, and moves from an angle of \(265^\circ\) to \(275^\circ\). If the pendulum is \(1\) metre in length, what is the minimum width the base of the clock must be? Round to the nearest hundredth of a metre.
Step 1: Draw a diagram.
Calculate the reference angle for both \(265^\circ\) and \(275^\circ\).
Calculate the reference angle for both \(265^\circ\) and \(275^\circ\).
\(\begin{align}
\theta _R &= \theta - 180^\circ \\
\theta _R &= 265^\circ - 180^\circ \\
\theta _R &= 85^\circ \\
\end{align}\)
\theta _R &= \theta - 180^\circ \\
\theta _R &= 265^\circ - 180^\circ \\
\theta _R &= 85^\circ \\
\end{align}\)
\(\begin{align}
\theta _R &= 360^\circ - \theta \\
\theta _R &= 360^\circ - 275^\circ \\
\theta _R &= 85^\circ \\
\end{align}\)
\theta _R &= 360^\circ - \theta \\
\theta _R &= 360^\circ - 275^\circ \\
\theta _R &= 85^\circ \\
\end{align}\)
Step 2: Determine the value of \(x\) using the reference angle.
\(\begin{align}
\theta_R &= 85^\circ \\
\rm{adj} &= \frac{x}{2} \\
\rm{hyp} &= 1 \rm \thinspace m\\
\end{align}\)
\(\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos 85^\circ &= \frac{{\left( {\frac{x}{2}} \right)}}{1} \\
\cos 85^\circ &= \frac{x}{2} \\
2\cos 85^\circ &= x \\
0.174... &= x \\
\end{align}\)
Step 3: Add the width of the pendulum weight.
The radius is \(5\) cm; therefore \(0.05\) m must be added to each side of the swing, or \(0.10\) m total.
\(\begin{align}
0.174... + 0.10 &= 0.274... \\
&\doteq 0.27{\rm{ m}} \\
\end{align}\)
The base of the clock must be at least \(0.27\) m wide.