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Q. 70

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 418

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Short Answer

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° amplitude when its lower

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

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

has decreased to 7.0°. Fortunately, the prisoner is freed with a

mere 30 s 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 ω=0.63rad/s

See the step by step solution

Step by Step Solution

Step 1: Given data

Data from given information

ϕo=10

The length of the pendulumL=6mThe radius of blade is rblade=1m

The lower edge is yo=3m

vy=1mm/s=0.001m/s= constant

After25 minutes, 1500s, the amplitude of the pendulum is 7.0 degree

Step 2:  Solution

To find the time constant of damped oscillation for the pendulum

ϕmax(t)=ϕoet2τ

t=1500s

ϕ=7=0.7ϕo

0.7ϕo=ϕoe1500s2τ

0.7=e1500s2τ

ln(0.7)=1500s2τ

τ=1500s2ln(0.7)=2102.75s

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

yf=yi+vyt

0=yi+vyt

yi=vyt

t=yivy

t=(3m)0.001m/s=3000s

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=2πImgL.

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

localid="1650084825110" I=12mr2

Substituting the values in equation

localid="1650084830138" T=2πmr22mgL

localid="1650084836858" T=2πr22gL

localid="1650084843529" T=2π(1m)229.8m/s2(6m)=0.58s/osc.

so localid="1650084849770" 2970s0.580cc/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)=2969.75 s

Step 5

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

E(t)=Eoe(-t/t)(3)

So, let t=2969.75 s. Observe that at this time, the energy in the system is entirely kinetic since ϕ=0 when the blade is above Bond. Thus, E(2969.75""s)=1/2mv2,where v is the speed that we want. Further, observe thatEo is entirely gravitational potential since the blade is released from rest. Thus, Eo=mgyo. Put the two equations in equation (3) and solve for v at t = 2969.75 s:

12mv2=mgyoe2969.75sT

v=2gyoe2969.75sτ

v=29.8m/s2(3m)e2999.75s2102.75s=3.78m/s.

Step 6

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

v=ωr

ω=vr

ω=3.78m/s6m=0.63rad/s

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