Angular Rotation

  • Thread starter Nanuven
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[SOLVED] Angular Rotation

Homework Statement



A cockroach of mass m lies on the rim of a uniform disk of mass 7.50 m that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with an angular velocity of 0.250 rad/s. Then the cockroach walks half way to the center of the disk.
(a) What then is the angular velocity of the cockroach-disk system?
_______ rad/s

(b) What is the ratio K/K0 of the new kinetic energy of the system to its initial kinetic energy?
______


(c) What accounts for the change in the kinetic energy?
centrifugal force
friction
cockroach does negative work on the disc
cockroach does positive work on the disc
gravity
centripetal force



The Attempt at a Solution



Ok so Rotational Inertia of the Disk at all times would be (1/2)MR^2
Then the Rotational Inertia of the Bug would be mR initially and then (1/2)mR finally.

Since momentum is conserved Iw = Iw

Therefore,

(1/2)MR^2(.25) + (m)(R)(.25) = (1/2)MR^2(w) + (1/2)mR(w)

I have the mass of the uniform disk so I can plug that into M giving me

3.75R^2(.25) + (m)(R)(.25) = 3.75R^2(w) + (1/2)mR(w)

Now I'm lost, can anyone point me in the right direction? Thanks
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi Nanuven,

The Attempt at a Solution



Ok so Rotational Inertia of the Disk at all times would be (1/2)MR^2
Then the Rotational Inertia of the Bug would be mR initially and then (1/2)mR finally.

Since momentum is conserved Iw = Iw

Therefore,

(1/2)MR^2(.25) + (m)(R)(.25) = (1/2)MR^2(w) + (1/2)mR(w)

I have the mass of the uniform disk so I can plug that into M giving me

3.75R^2(.25) + (m)(R)(.25) = 3.75R^2(w) + (1/2)mR(w)

Now I'm lost, can anyone point me in the right direction? Thanks

The rotational inertia of the bug would be mR^2. After you have that, what would be the final value of its rotational inertia?
 

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