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3 years ago

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.4 m/s . What upward force must a branch provide to support the swinging gibbon?

1 answer:

Katarina [22]3 years ago

7 0

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

$F_{branch} + F_{gravity}=F_{centripetal}$

force of gravity = - mg

$F_{branch}=F_{centripetal}-F_{gravity}$

= $\dfrac{mv^2}{r} + mg$

= $m(\dfrac{3.4^2}{0.6} +9.8)$

=9 x 29.067

= 261.6 N

upward force acting = 261.6 N

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(a) 2.75 rev/min

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When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

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$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

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