Angular Rotation - Find Angular Velocity & Kinetic Energy Ratio

[Nanuven]

[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

 

[alphysicist]

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?

 

[Moetasim]

!





To find the new angular velocity of the cockroach-disk system, we can use the conservation of angular momentum. The initial angular momentum of the system is given by (1/2)MR^2(0.250) + (m)(R)(0.250) = 3.75R^2(0.250) + (1/2)mR(0.250). After the cockroach walks halfway to the center, the new angular momentum is given by (1/2)MR^2(w) + (1/2)mR(w). Setting these two equal to each other and solving for w, we get w = 0.375 rad/s.

To find the ratio of the new kinetic energy to the initial kinetic energy, we can use the formula K/K0 = (w/w0)^2, where w0 is the initial angular velocity and w is the new angular velocity. Plugging in the values, we get K/K0 = (0.375/0.250)^2 = 2.25.

The change in kinetic energy is due to the cockroach doing positive work on the disk. As the cockroach walks towards the center, it decreases the moment of inertia of the system, causing the angular velocity to increase. This results in an increase in kinetic energy. There is also a decrease in potential energy due to the decrease in distance between the cockroach and the center of rotation. This change in potential energy is converted to kinetic energy, resulting in an overall increase in the kinetic energy of the system.