If a buggy and a sphere of equal masses begin rolling side-by-side down a hill, which will reach the bottom of the hill first ( neglecting wind resistance ) ?
For simplicity, we should pretend that the energy transfer in this problem is 100% efficient: no losses occur due to sound, random vibrations, or wind resistance. All of the potential energy that would be possessed by a body sitting atop a hill will be assumed to be 100% converted to kinetic energy ( the energy of motion ).
Ei ( Initial energy ) = Ef ( Final energy ). If this statement is true, we must determine how to mathematically quantify the energy states of each body ( 1 ) sitting stationary atop a hill, and ( 2 ) moving at full speed after being allowed to roll downhill. Measurements can be made to prove that potential energy is quantified by the expression mgh, where m = mass ( kilograms ), g = gravitational constant of acceleration ( meters / second^2 ), and h = height ( meters ) in regard to some vantage point.
Thus far, we know that the potential energy ( mgh ) will be converted into the energy of motion. Two forms of motion will be of concern here: linear and rotational motion. Each body will be moving linearly, and they will also be in rotation as they reach the bottom of the hill.
Understanding the equation used to represent rotational energy will be easier to understand if we draw a circle within a larger circle. If both circles have a common center, a line can be drawn from the center until it passes the outer edge of the smaller ( inside circle ) and ends upon reaching the edge of the outer ( larger ) circle.
Starting again at the common center, another line can be similarly drawn such that a ” slice of pie ” is taken out of both circles. The length ” l ” of the edge of the inner ” slice of pie ” will be smaller than the length of the outermost edge contained within the common angle. However, the radial distance r2 defining the second larger ” slice ” will be greater than r1. SINCE THE ANGLE BETWEEN THESE TWO SUPERIMPOSED SLICES OF CIRCLE ARE THE SAME, THE RATIOS OF THE RADII OF R1 TO L1 WILL BE THE SAME AS THAT FOR R2 TO L2. From here on out, the angle referred to will be called ” theta “. Theta is defined as a length ( l ) along a circle’s outer edge divided by the circle’s radius ( r ) for any circle of any size.
Now, let’s continue with rotational motion. ( l / r ) = theta. Therefore, l = ( r )( theta ). Recall that ” l ” describes an outside length along the curve of a circle as determined by some angle. Since velocity ” v ” is defined as distance ” l ” divided by time ” t ” ( v = l/t ), we must divide ” l ” by time on the left-hand side of the equation to get ( v ), and we must divide ” theta ” on the right-hand side of the equation to get the rotational equivalent of velocity ( angular velocity in radians / second ) = ( theta / t ) = omega ( w ). Thus, our new equation, v = ( r )( w ), relates linear velocity with the angular ( rotational ) velocity of the circle.
If we now square all terms, we get v^2 = ( r^2 )( w^2 ). Recall that centripetal acceleration is defined by ( v^2 / r ). Therefore, if we divide each term by ” r “, we have an expression for centripetal acceleration. a = ( r )( w^2 ).
Force ” F” is equal to mass ” m ” times acceleration ” a “. F = ( m )( a ). In the equation above, F = ( m )( a ) = ( m )( r )( w^2 ).
Finally, energy is defined by force in Newtons ( N ) times distance. Think in terms of an arm pulling on a wrench. The wrench’s length is like the radius of one of the aforementioned circles, and the force is like an arm pulling the wrench perpendicularly. Work ( energy ) = ( F )( r ). We have already derived an equation for force, so we now need to multiply it by ” r “. We will now have an expression for the angular energy gained by a rotating sphere. ( F )( r ) = ( m )( r^2 )( w^2 ). The average energy ( as is the case with kinetic energy ) is multiplied by a factor of ( 1 / 2 ). THUS, ENERGY OF ROTATION = ( 1 / 2 )( m )( r^2 )( w^2 ).
NOTE: The term ( m )( r^2 ) is of special importance; it quantifies the resistance a body has ( moving or sitting still ) to having its current state of motion changed. The expression ( m)( r^2 ) is usually denoted as ( I ), and it is referred to as the ” Moment of Inertia “. When a figure skater pulls her/his arms inward, s/he is reducing his/her effective radius ” r ” in terms of rotational inertia. Since angular momentum must be conserved, the reduction of inertia ” I ” is observed with an increase in angular velocity ” w ” in the skater.
So, we now have an expression for rotational energy: E = ( 1 / 2 )( I )( w^2 ). Each body began with potential energy ( mgh ). In addition to rotational energy, each body will also have kinetic energy ( 1 / 2 )( m ) v^2 ). Let’s put this into equation form:
mgh = ( 1 / 2 )( m )( r^2 )( w^2 ) + ( 1 / 2 )( m )( v^2 ). We can divide the ” m ” out of each equation to get gh = ( 1 / 2 )( r^2 )( w^2 ) + ( 1 /2 )( v^2 ).
Recall that w = ( v / r ), and thus, w^2 = ( v^2 / r^2 ). Let’s simplify the above equation by substituting for w^2.
gh = ( 1 / 2 )( r^2 )( v^2 / r^2 ) + ( 1 / 2 )( v^2 ). Simplifying this expression gives us gh = ( 1 / 2 )( v^2 ) + ( 1 / 2 )( v^2 ). ONE HALF OF v^2 PLUS ANOTHER HALF OF v^2 gives us a whole unite of v^2. gh = v^2. Therefore, the final velocity of EITHER of these moving bodies will be the square root of gh. SQRT ( gh ) = v.
Which body, however, will reach the bottom of the hill first? Well, the sphere had to overcome more rotational inertia ( review the above derivation again, paying attention to ( m )( r^2 ) as being a measure of inertia ” I “. The wheels of the buggy have smaller radii, and their rotational inertial energy barrier is easier to overcome. Therefore, the buggy will reach the bottom of the hill first.
RENAISSANCE PHYSICS 