5 uo pesn George of the with mass of kg, swings on light th Lof t 4 so m) from a

Question

5 uo pesn George of the with mass of kg, swings on light th Lof t 4 so m) from a stationary tree A second vine of equal length hangs from branch. swings the direction the same point, and a goriva a 120.o from rest the same that one the primates start is afraid on it. Both vines are of their swings. Each together. How rge and the gorilla meet at the lowest point They swing upward vine wil break, so they grab each other and hang on. high will they swing up? rod is released fro 8. A uniform rod is 2.5 m long. The rod pivoted about a frictionless pin through one end. The the is rest at its horizontal position and allowed to fall. angular acceleration al and a. pts/As the rod swings down, what happens to the magnitude of the one) magnitude the angular velocity Iwi? Hint: set up equations, don't just guessij the of /al increases, lol decreases li. Ial decreases, Iwl increases ii. /al remains constant, Iwl increases iv. /al decreases, Iw/ decreases v /al remains constant, Iwl remains constant at an angle of S3 below 15 pts) What is the angular acceleration of the rod at the instant the rod is horizontal? pts] What is the angular velocity at that time? rm solid disk starts from rest and rolls down a 40.0 incline without The disk has a mass M of 5.00 kg, and a radius R of 3.50 cm. If the of the incline is 1.50 m, What is the linear speed of the disk as the bottom of the incline?

Explanation / Answer

at any position theta with the horizontal

net torque = I*alpha

m*g*L/2*costheta = I*alpha

alpha = m*g*L/2*costheta/I

as the rod rotates theta value increases ang alpha decreases

the rod have angular accelration so the angular speed increases

OPTION ii

alpha decreases w increases

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alpha = m*g*L/2*costheta/I

I = (1/3)*m*L^2

alpha = m*g*L/2*costheta/(1/3)*m*L^2

alpha = (3/2)*g*costheta/L

alpha = (3/2)*9.8*cos53/2.5 = 3.54 rad/s^2

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(c)

In horizontal position

initial energy Ei = m*g*L/2

final energy Ef = m*g*L/2*(1-sintheta) + (1/2)*I*w^2

from energy conservation

Ef = Ei

m*g*L/2*(1-sintheta) + (1/2)*I*w^2 = m*g*L/2

(1/2)*I*w^2 = m*g*L/2*sintheta

(1/2)*(1/3)*m*L^2*w^2 = m*g*L/2*sintheta

(1/3)*L*w^2 = g*sintheta

(1/3)*2.5*w^2 = 9.8*sin53

w = 3.1 rad/s

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