Part A When you release the bola, where ls the center of maas of the boila? Use

Question

Part A When you release the bola, where ls the center of maas of the boila? Use the coordinate system dsplayed In the figure. Enter the x-position and y-position of the center of mass below. Separate the x pceition and y-position with a comma. Express your answer in meters and use 3 significant figures.) x-position, y position meters, meters Submit Part B In order for the center of the bola to hit the topmost rung of the ladder, how far away from the ladder in the x-direction does the centar of the bola need to be when you release the bola? Express your answer in meters and use 3 significant figures Submit

Explanation / Answer

part A:

the lower mass is at (0,h-l)=(0,0.48) m

the upper mass is at (0,h)=(0,0.8) m

the center of mass of the rope is at (0,(h-l)+(l/2))=(0,0.64) m

then center of mass of the entire system is at:

(45*(0,0.48)+45*(0,0.8)+10*(0,0.64))/(45+45+10)

=(0,0.64) m

part B:

acceleration of the system=

=(2.688,4.057)/total mass

=(26.88,40.57) m/s^2

velocity at the end of 0.13 seconds=initial velocity+acceleration*time

=(0,0)+(26.88,40.57)*0.13

=(3.4944,5.2741) m/s

for the center to reach the top rung of the stand,

vertical height covered=1-0.64=0.36 m

initial vertical speed=5.2741 m/s

acceleration=-9.8 m/s^2

let time taken be t seconds

then using the formula:

displacement=initial velocity*time+0.5*acceleration*time^2

==>0.36=5.2741*t-0.5*9.8*t^2

solving for t , we get

t=1 seconds or t=0.0732 seconds

discarding the lower value, taking t=1 seconds

horizontal distance=horizontal speed*time

=3.4944 m

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