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Q. 70

Expert-verifiedFound in: Page 418

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A captive James Bond is strapped to a table beneath a huge

pendulum made of a $2.0-m$-diameter uniform circular metal blade

rigidly attached, at its top edge, to a $6.0m$-long, massless rod.

The pendulum is set swinging with a $10\xb0$ amplitude when its lower

edge is $3.0m$ above the prisoner, then the table slowly starts ascending

at $1.0mm/s$. After $25$ minutes, the pendulum’s amplitude

has decreased to $7.0\xb0$. Fortunately, the prisoner is freed with a

mere $30s$ to spare. What was the speed of the lower edge of the

blade 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 $\omega =0.63\mathrm{rad}/\mathrm{s}$

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 $7.0degree$

To find the time constant of damped oscillation for the pendulum

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

$t=1500s$

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

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

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

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

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

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

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

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

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

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

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

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

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

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" $(5120.25osc)(0.58s/osc)=2969.75s$

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

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

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, $E(2969.75""s)=1/2m{v}^{2},$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 $t=2969.75s:$

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

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

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

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

$v=\omega r$

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

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

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