Conserving Angular Momentum: Solving a Cockroach Problem

[Tanya Sharma]

Homework Statement

A cockroach with mass m rides on a disk of mass 6.00m and radius R .The disk rotates like a merry go round around its central axis at angular speed ω_i=1.50rad/s.The cockroach is initially at radius r=.800R,but when it crawls out to the rim of the disk .Treat the cockroach as the particle.What then is the angular speed?

Using the conservation of angular momentum the correct answer is obtained.But I am not clear why can't we use conservation of energy in this problem since no external forces are acting on the system of disk and cockroach.

Homework Equations

The Attempt at a Solution

 

[rcgldr]

I am not clear why can't we use conservation of energy.

In this case, the cockroach does "negative" work as it crawls towards the outside of the disk. The "negative" work done is the intergral of the centripetal force (as a function of r) x Δr (change in radius) from .8 R to 1.0 R.

 

[Tanya Sharma]

rcgldr...Thank you for the response...Why centripetal force? Please can you elaborate...

 

[rcgldr]

rcgldr...Thank you for the response...Why centripetal force? Please can you elaborate...

Work = force x distance. In this case the force is the centripetal force, and the distance is radial, from .8 R to 1.0 R. The force varies as r goes from .8 R to 1.0 R, so using an integral would be the normal way to determine the work done. Since the movement is outwards, the work done is "negative". During it's movement outwards, the path of the cockroach is a spiral, so there's a component of opposing force in the direction of the spiral path, decreasing the speed.

grinding out the math:

let r = position of bug from center

define unit of mass to be "1 bug", so that m = 1

define unit of length to be R so that R = 1

define unit of angular velocity to be 1 radian / second.

Angular momentum is conserved:

IB = inertia of bug = m r² = r² {since m = 1}
ID = inertia of disk = 1/2 6 m R² = 3 {since m = 1 and R = 1}

Initial state:

r = .8
w(r) = w(.8) = 1.5
L(.8) = (IB + ID) w(.8) = (.64 + 3) 1.5 = 5.46

Angular momentum is conserved:

L(r) = (IB + ID) w(r) = 5.46
L(r) = (r² + 3) w(r) = 5.46
w(r) = 5.46 / (r² + 3)

w(1) = 5.46 / (1 + 3) = 1.365

energy versus r = e(r)

e(r) = 1/2 (IB + ID) w(r)²
e(r) = 1/2 (r² + 3) (5.46 / (r² + 3))²
e(r) = 14.9058 / (r² + 3)

e(0.8) = 14.9058 / (.64 + 3) = 4.09500
e(1.0) = 14.9058 / (1.0 + 3) = 3.72645

energy change = e(1) - e(.8) = -0.36855

velocity of bug versus r = v(r)

v(r) = w(r) r = 5.46 r / (r² + 3)

force on bug = f(r)

f(r) = - m v(r)² / r = -1 (5.46 r / (r² + 3))² / r
f(r) = -29.8116 r / (r² + 3)²

work done by bug moving from .8 to 1 R

w = \int_{.8}^1 - 29.8116 \ r \ dr / (r^2 + 3)^2
w = \left. 14.9058 / (r^2 + 3) \right]_{.8}^{1}
(note this matches the formula for energy versus r)

w = 14.9058 (1/(1+3) - 1/(.64+3)) = -0.36855

so the negative work done by the bug matches the energy change and accounts for the loss in energy.

 

[Tanya Sharma]

rcgldr...Thank you very much...

f(r) = - v(r)² / r

Plz can you explain it ...

Thanks

 

[rcgldr]

f(r) = - m v(r)² / r, since m = 1, then f(r) = -v(r)² / r

Plz can you explain it.

That's the centripetal force on the bug as a function of r (the radius). I used -m v(r)² / r since the direction of centripetal force is inwards, and I choose the unit of mass to be "1 bug", so m = 1.

For a similar example, imagine that you are on a merry-go-round, and that it's mass is 6 times yours and that it's inertia is the same as a uniform disk with same radius. This time, imagine that you start on the outside of the merry-go-round, and pull yourself inwards. Angular momentun is conserved, and the work you exert in pulling yourself inwards increases the energy by the amount of work done. Using the original problem scenario, when you're at the outside edge, R, the rate of rotation is 1.365 rad / sec. At .8 R, it's 1.5 rad / sec, and when you reach the middle (0 R), it's 5.46 / (0 + 3) = 1.82 rad / sec, and the energy is 4.9686.

For another example, imagine a puck whirling around on a frictionless table, attached to a string that is beign pulled at the center through a hole. All of the angular momentum and energy is in the puck, so the math is different. The math is shown in post #3 of this old thread:

https://www.physicsforums.com/showthread.php?t=328121

 

[Tanya Sharma]

Thanks for the wonderful explanation...

That's the centripetal force on the bug as a function of r (the radius). I used -v(r)² / r since the direction of centripetal force is inwards.

Somehow one thing is not very clear...How have you written this force as a function of r...I want to understand why f(r) = -v(r)² / r ?

 

[rcgldr]

Somehow one thing is not very clear...How have you written this force as a function of r...I want to understand why f(r) = -v(r)² / r ?

Centripetal force = m v² / r, and I defined the unit of mass to be "1 bug" so that m = 1. In this case the velocity v can be expressed as a function of the radius of the bug's current location on the disk, and I use v(r) to mean v as a function of r. Linear velocity for a point at r = ω r, where ω is the angular velocity. Since ω can also be expressed as a function of r, I used ω(r).

 

[Tanya Sharma]

rcgldr... Centripetal force is \frac{mv^2}{r}...why is m not included?

 

[ehild]

But I am not clear why can't we use conservation of energy in this problem since no external forces are acting on the system of disk and cockroach.

No external force does not mean conservation of energy in case of a system. Think of two colliding bodies -there is no external force, but the energy conserves only in the very special case of elastic collision. If the internal forces, the forces of interaction, are not conservative the energy of a system of bodies does not conserve.

ehild

 

[rcgldr]

rcgldr... Centripetal force is \frac{mv^2}{r}...why is m not included?

Sorry, I didn't understand what your were asking. The mass of the bug was given as m, while the mass of the disk was given as 6 m. I just defined the unit of mass to be "1 bug", so m = 1, and unit of length to be "R", so that R = 1 for this problem to keep the math simple. I also left out other units, such as radians / second, and the units for r. Centripetal accelertaion is v² / r, while centripetal force = m v² / r. Since m = "1 bug", I left out the unit. I updated my previous posts to fix this, starting with post #4. Again, sorry for leaving out the mass term.

 

[Tanya Sharma]

No external force does not mean conservation of energy in case of a system. Think of two colliding bodies -there is no external force, but the energy conserves only in the very special case of elastic collision. If the internal forces, the forces of interaction, are not conservative the energy of a system of bodies does not conserve.

ehild

In elastic collisions the bodies do exert forces ,say body 1 exerts force f1 on body 2 ... This force is doing work as the bodies may move during collision ...Why is kinetic energy being conserved here even though work is been done?

 

[rcgldr]

In elastic collisions the bodies do exert forces ,say body 1 exerts force f1 on body 2

Consider the two bodies as a closed system with no external forces. The only forces are internal to this closed two body system, and assuming elastic collisions, no energy is lost, so the total energy of the system remains constant. During the collision, some or all of the energy is stored as potential energy related to deformation (compression) of the two bodies.

 

[Tanya Sharma]

rcgldr...how is the original bug problem different from elastic collision in the sense both are two body system having no external forces and one body is doing work ...work is being done in collisions also...isnt it ?

 

[ehild]

In elastic collisions the bodies do exert forces ,say body 1 exerts force f1 on body 2 ... This force is doing work as the bodies may move during collision ...Why is kinetic energy being conserved here even though work is been done?

Elastic collision means the transformation of kinetic energy of the bodies into some elastic energy for a short time, and then transformation back to kinetic energy.

ehild

 

[rcgldr]

rcgldr...how is the original bug problem different from elastic collision in the sense both are two body system having no external forces and one body is doing work ...work is being done in collisions also...isnt it ?

Although the bug problem involves an internal force, ultimately the negative work done by the bug is being converted into heat. If the bug were moving inwards, then potential energy within the bug is being converted into kinetic energy. A similar situation occurs when an ice skater pulls in his/her arms during a spin. The skater uses chemical potential energy to perform "postiive" work. The bug could be replaced by some type of mechanical device that converts the negative work into potential energy within the mechanical bug. Assuming that the mechanical bug was 100% efficient, then the total energy (potential + kinetic) of the mechanical bug + disk closed system would remain constant.

For the elastic two body system, during the period of collision, the kinetic energy of the system is converted into potential energy within the two bodies, as if they were springs. Since the collision is elastic, once they separate, all of that potential energy is retuned back as kinetic energy.

 

[Tanya Sharma]

ehild...rcgldr...thank you very much for your time and energy...wonderful insight in conservation of angular momentum and energy

 

[rcgldr]

ehild...rcgldr...thank you very much for your time and energy...wonderful insight in conservation of angular momentum and energy

The issue with the bug problem is that energy is lost, since a real bug can't increase it's internal potential energy by doing negative work. Again, sorry for not originally including how I defined the units of mass and length in post #4 to keep the equations simple (it's fixed now).