ENERGY AND MOMENTUM: Translational and Rotational Kinetic Energy

When determining the final kinetic energy ( KE ) of falling objects, we need not ( in theory ) concern ourselves with anything other than the linear pathway traveled to the earth’s surface. To the contrary, an object that rolls top to bottom down an incline will gain both linear ( KE ) and rotational ( KE ) as well. This requires us to assume that the net sum of mass contained by an object is concentrated in its center of mass ( cm ). As far as rotational ( KE ) is concerned, the mass equivalent of the system is the moment of inertia ( I ):

I = mr²

One notable example of the moment of inertia at play can be observed when watching figure skaters skate. Whenever a skater wants to increase the rotational speed ( ⍵ ) of her spin, she pulls her arms inward to effectively decrease ( I ). With her arms tucked inward, the energy vectors that helped propel her extended arms round and round are now positioned closer to her body. This ( KE ) is added to the ( KE ) vectors that were already close to her trunk beforehand, and her lessened inertia is accompanied by an observable increase in angular speed. The angular speed of an object is derived as follows:

θ = ( l / r )

A look at an angle measurement across two circles ( one within the other ) will reduce the ambiguity of this equation:

The inner circle has a radius ( r₁ ), and the larger circle that surrounds it has a radius ( r₂ ). Regardless of how large or small a circle is, that angle that describes some displacement equal to ( l ) along the circle’s edge will be the same. With the case being made, we can now derive an expression that relates a linear speed vector ( v ) of a displaced point along ( l ) with the angular speed ( ⍵ ) that describes this motion:

l = rθ

The angle theta ( θ ) describes the number of radians ( l ) that an imaginary particle has traveled along a circle’s outer surface. A radian ( 1 rad ) is equal in length to a circle’s radius ( r ), and the diameter ( D = 2𝞹r  ) of a circle contains a multiple of ( 2𝛑 ) radians within it.

Now, getting back to our derivation. Since ( l ) is a measure of distance, with ( θ ) describing the magnitude of an angle that subtends or coincides with a displacement along ( l ), dividing both sides by time ( t ) measured in seconds yields an expression that relates the velocity of such motion to angular speed measured in radians per second:

v = rω

Our ultimate aim is to express angular speed in terms of linear speed and velocity:

ω = ( v / r )

We can continue our expansion of the velocity term until we arrive at an expression for rotational ( KE ):

v² = r²ω²

mv² = mr²ω²

KE = ½ mv²

KE_r = ½ mv² = ½ mr²ω²

I = mr²

KE_r = ½ Iω²

We are now ready to take a look at an object that is about to roll down an incline that is ( θ ) radians above and diagonal to the surface below it:

It is important to note that the moment of inertia will differ from object to object!!! For this reason, care must be taken to use ( m-o-i ) values that fit the system at hand. In terms of ( PE ) and ( KE ) for the system above, the following equation will be expanded:

KE_i + PE_i = KE_f + PE_f

The initial ( KE ) of the system is zero with the final ( PE ) of the system being equal to zero itself; thus, the equation can be simplified further:

PE_i = KE_f

Since the final ( KE ) of the system is both linear and rotational in nature, and the ( KE_f ) term must reflect this:

PE_i = KE _linear + KE _rotational

The ( PE ) term is the now-familiar ( right ??? 🙂 ) term that relates weight ( F_w = mg ) to the vertical distance ( Δy ) of displacement:

ΔPE = mgΔy

We now have all of the terms needed to solve introductory-level problems in which an object rolls down an incline until it reaches some final velocity, with velocity being related to the final rotational and linear ( KE ):

mgΔy = ½ mv² + ½ Iω²

Q: Assume the object on the incline above is a disc with the following moment of inertia value:

I = ½ mr²

What is the final speed of the disc’s center of mass if it rolls without slippage from a height ( Δy ) to the bottom of the incline?

A: PE_i = KE _linear + KE _rotational

mgΔy = ½ mv² + ½ Iω²

mgΔy = ½ mv² + ½ ( ½ mr² )ω²

mgΔy = ½ mv² + ¼ mr²ω²

gΔy = ½ v² + ¼ r²ω²

ω = ( v / r )

ω² = ( v² / r² )

mgΔy = ½ mv² + ( ¼ mr² )( v² / r² )

mgΔy = ½ mv² + ¼ mv²

mgΔy = ²/₄ mv² + ¼ mv²

gΔy = ²/₄ v² + ¼ v²

gΔy = ¾ v²

⁴/₃ gΔy = v²

v = √ ⁴/₃ gΔy 

As is the case with single objects falling from an arbitrary height ( Δy ) to a fluid of unknown density ( ρ ) flowing through a spigot of a container of height ( Δy ), the mass cancels out of the equation. Mass is a measure of inertia, and although a greater amount of mass moving downward from some height will have more inertia, the loss of energy of each comparable unit of mass between one system and another will be one and the same, as should be the case.