Bob the monkey (10 kg) stands on a small platform (3 kg), which is attached by m

Question

Bob the monkey (10 kg) stands on a small platform (3 kg), which is attached by massless rods (R = 4 m) to a frictionless pivot, and initially at rest. Bob would like to get the banana shown, which he could reach-- if only the platform were rotated a half-circle from its current location. Being a smart monkey, he starts spinning around his center of mass on the platform... and soon has his snack. What is the size of this entire system's angular momentum: when it is in the state described above: kg middot m^2/s if Bob starts spinning around his own COM: kg middot m^2/s (Make sure you understand why.) What is Bob's rotational inertia about his own center of mass? (You may approximate him as a sphere of radius 0.5 m) kg middot m^2 What is the rotational inertia of the system around the frictionless pivot? (Remember that Bob also moves with the platform. In this calculation, treat both him and the platform as point masses.) kg middot m^2 If Bob spins himself around his own COM with a constant* angular speed of 70 rad/s, *Please neglect his acceleration from/deceleration to rest. how do Bob+platform rotate about the frictionless pivot? rotation direction of Bob around his COM how long does it take for Bob to get his snack? s

Explanation / Answer

given that

mb = 10 kg

mp = 3 kg

R = 4 m

(b)

we know that

moment inertia of solid sphere is

Io = (2/5)*M*r^2

Io = (2/5)*(10 + 3)*(0.5)^2

lo = 1.3 kg*m^2

(b)

using parallel axis theorem

I' = Io + M*R^2

I' = 1.3 + (10 + 3)*4^2

l' = 209.3 kg*m^2

(d)

using conservation of angular momentum;

Li = Lf

0 = Io*w + I'*w'

0 = 1.3*70 + 209.8*w'

w' = -0.434 rad/s

rotation direction of bob = clockwise

(e)

theta = w'*t

bob+plateform will rotate 180 deg to get his snack

theta = 180 deg = 3.14 rad

t = 3.14 / 0.434

t = 7.23 s

(a)

because the monkey is initialy at rest

angular momentum = 0

let angular momentum after bob starts spinning aroun his owm COM is L'

L' = l*w + I*w' = 1.3*70 + 209.3*0.434

L' = 181.83 kg*m^2/s

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