A 90.5-kg horizontal circular platform rotates freely with no friction about its

Question

A 90.5-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.71 rad/s. A monkey drops a 9.89-kg bunch of bananas vertically onto the platform. They hit the platform at 4/5 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 22.1 kg, drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.55 m.

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.837 m and 1.25 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.57 rad/s in 9.47 s?

Explanation / Answer

Given

Mass of the Disk, Md = 90.5 kg,

mass of bananas, Mb = 9.89 kg,

mass of the monkey, Mm =22.1 kg

Initial angular velocity ,wi =1.71 rad/s

Moment of Inertia of the disk axis passing through the centre & perpendicular to the plane of disk =Md(R^2)/2,

Moment of inertia of bunch of bananas about axis of rotation = Mb((4/5)R)^2,

Moment of inertia of the monkey about axis of rotation = Mm(R)^2;

Apply

Conservation of angular momentum,

Intial total momentum = Final total momentum

==> Li = Lf

==> I(disk).wi= [I(disk) + I(ban)+ I(mon)].wf

==> {Md(R^2)/2}x1.71 ={ Md(R^2)/2 + Mb((4/5)R)^2 + Mm(R)^2}x w

==> wf =1.05 rad/sec. Is the final angular velocity.

b)

moment of inertia of the rod rotating about one end, I =(ML^2)/3;

we have

wf = wi + (alpha)t

==> 6.57= 0 + (alpha)x 9.47

==> (alpha)=0.693 rad/s^2 ;

Net force will be tangential to circle formed by the rotation

its components are

F sinQ & F cosQ

these are centripetal & centrifugal forces respectively.

apply

Angular momentum = R x P & Angular momentum = I (w)

==> I (w) = R x P

==> I (wf - wi) = R x F net (tf - ti)

==> (ML^2)/3 (6.57 - 0) = L {sqrt 2}F (9.47 -0)

==> Fnet = 0.17 N

therefore,

F centripetal = F net sinQ

= Fnet sin45

=0.12 N

F centrifugal = F net cosQ

= Fnet cos 45

=0.12 N

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