Question

A 1.7-kg monkey wrench is pivoted 0.20 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.91 s.

If the wrench is initially displaced 0.252 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position, in rad/s?

Answer 1

Time period of physical pendulum is given by,

T = 2*pi*sqrt(I/(m*g*d))

here, m = mass of monkey = 1.7 kg

g = surface gravity = 9.81 m/s^2

d = length of wrench = 0.20 m

I = moment of inertia = ??

T = time period = 0.91 s

So,

0.91 = 2*pi*sqrt(I/(1.7*9.81*0.20))

I = 1.7*9.81*0.20*(0.91/(2*pi))^2

I = 0.07 kg*m^2

Now using energy conservation between initial position and equilibrium position:

KEi + PEi = KEf + PEf

here, KEi = initial kinetic energy = 0

PEi = initial gravitational potential energy = m*g*h

KEf = final kinetic energy = 0.5*I*w^2

PEf = final gravitational potential energy = 0

h = initial height of monkey from equilibrium position = d - d*cos

given, =  initially angular displacement = 0.252 rad

So,

0 + m*g*d*(1 - cos) = 0.5*I*w^2 + 0

w = angular speed of the wrench as it passes through the equilibrium position = sqrt(2*m*g*d*(1 - cos)/I)

w = sqrt(2*1.7*9.81*0.20*(1 - cos(0.252))/0.07)

w = 1.735 rad/s

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