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Q. 70

Expert-verified
Found in: Page 418

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# A captive James Bond is strapped to a table beneath a hugependulum made of a $2.0-m$-diameter uniform circular metal bladerigidly attached, at its top edge, to a $6.0m$-long, massless rod.The pendulum is set swinging with a $10°$ amplitude when its loweredge is $3.0m$ above the prisoner, then the table slowly starts ascendingat $1.0mm/s$. After $25$ minutes, the pendulum’s amplitudehas decreased to $7.0°$. Fortunately, the prisoner is freed with amere $30s$ to spare. What was the speed of the lower edge of theblade as it passed over him for the last time?

The speed of the lower edge of the blade as it passed over him for the last time is $v=3.78m/s$ and $\omega =0.63\mathrm{rad}/\mathrm{s}$

See the step by step solution

## Step 1: Given data

Data from given information

${\varphi }_{o}={10}^{\circ }$

The length of the pendulum$L=6\mathrm{m}$The radius of blade is ${r}_{blade}=1\mathrm{m}$

The lower edge is ${y}_{o}=3\mathrm{m}$

${v}_{y}=1\mathrm{mm}/\mathrm{s}=0.001\mathrm{m}/\mathrm{s}=\text{constant}$

After$25$ minutes, $1500s$, the amplitude of the pendulum is $7.0degree$

## Step 2:  Solution

To find the time constant of damped oscillation for the pendulum

${\varphi }_{max}\left(t\right)={\varphi }_{o}{e}^{-\frac{t}{2\tau }}$

$t=1500s$

$\varphi ={7}^{\circ }=0.7{\varphi }_{o}$

$0.7{\varphi }_{o}={\varphi }_{o}{e}^{-\frac{1500\mathrm{s}}{2\tau }}$

$0.7={e}^{-\frac{1500\mathrm{s}}{2\tau }}$

$\mathrm{ln}\left(0.7\right)=-\frac{1500\mathrm{s}}{2\tau }$

$\tau =-\frac{1500\mathrm{s}}{2\mathrm{ln}\left(0.7\right)}=2102.75\mathrm{s}$

## Step 3:  To find t

Calculate the time at which the bond is freed. observe that it takes place $30s$ before $y=0m$

Then, the velocity in $y$ is constant

${y}_{f}={y}_{i}+{v}_{y}t$

$0={y}_{i}+{v}_{y}t$

$-{y}_{i}={v}_{y}t$

$t=\frac{-{y}_{i}}{{v}_{y}}$

$t=\frac{-\left(-3\mathrm{m}\right)}{0.001\mathrm{m}/\mathrm{s}}=3000\mathrm{s}$

Bond is saved with $30s$ to spare, then as we know $t=2970s$

## Step 4:

Now, we have to determine how many oscillation presented before bond is saved. For that, we have to find the period of oscillation of pendulum

$T=2\pi \sqrt{\frac{I}{mgL}}.$

Where $I$ is the moment of inertia of system. Therefore, the moment of inertia of system is

localid="1650084825110" $I=\frac{1}{2}m{r}^{2}$

Substituting the values in equation

localid="1650084830138" $T=2\pi \sqrt{\frac{m{r}^{2}}{2mgL}}$

localid="1650084836858" $T=2\pi \sqrt{\frac{{r}^{2}}{2gL}}$

localid="1650084843529" $T=2\pi \sqrt{\frac{\left(1\mathrm{m}{\right)}^{2}}{2\left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right)\left(6\mathrm{m}\right)}}=0.58\mathrm{s}/\mathrm{osc}.$

so localid="1650084849770" $\frac{2970\mathrm{s}}{0.580\mathrm{cc}/\mathrm{s}}=5120.69$ oscillation before bond is freed.

since, the previous time the blade is above bond would be after localid="1650084855792" $5120.25$ oscillations or afterlocalid="1650084863032" $\left(5120.25osc\right)\left(0.58s/osc\right)=2969.75s$

## Step 5

Then, we need to find the speed of the blade at $t=2969.75s$

$E\left(t\right)={E}_{o}{e}^{\left(}-t/t\right)\cdot \left(3\right)$

So, let $t=2969.75s$. Observe that at this time, the energy in the system is entirely kinetic since $\varphi =0$ when the blade is above Bond. Thus, $E\left(2969.75""s\right)=1/2m{v}^{2},$where v is the speed that we want. Further, observe that${E}_{o}$ is entirely gravitational potential since the blade is released from rest. Thus, ${E}_{o}=mg{y}_{o}$. Put the two equations in equation $\left(3\right)$ and solve for v at $t=2969.75s:$

$\frac{1}{2}m{v}^{2}=mg{y}_{o}{e}^{-\frac{2969.75\mathrm{s}}{T}}$

$v=\sqrt{2g{y}_{o}{e}^{-\frac{2969.75\mathrm{s}}{\tau }}}$

$v=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right)\left(3\mathrm{m}\right){e}^{-\frac{2999.75\mathrm{s}}{2102.75\mathrm{s}}}}=3.78\mathrm{m}/\mathrm{s}.$

## Step 6

The linear speed of it. The angular speed of the blade can be obtained by

$v=\omega r$

$\omega =\frac{v}{r}$

$\omega =\frac{3.78\mathrm{m}/\mathrm{s}}{6\mathrm{m}}=0.63\mathrm{rad}/\mathrm{s}$