Q: A student sits atop a freely rotating stool holding two dumbbells, each of which has a mass of 3.09 kg. When the student’s arms are extended horizontally outward, the dumbbells are 0.99 m from the axis of rotation. There are 180 degrees of separation between the extended arms. The student rotates with an angular speed of 0.749 rad/s. The moment of inertia ( I ) of the student + stool is 2.52 kg·m2. As the stool spins, the student pulls the dumbbells inward to a position located 0.298 m from the axis of rotation.
(a) What is the angular speed ( ⍵ ) of the system after the student’s arms have been moved inward?
(b) What is the kinetic energy of the rotating system before and after the dumbbells have been pulled inward?
A: This question regards the conservation of angular momentum. Linear momentum ( p = mv ) is oftentimes described as being a ” quantity of motion “. For example, let’s say that a cannon has a mass of 250 kg. The cannonball it fires has a mass of 5.0 kg. When the cannon is fired, equal and opposite forces will exert themselves upon the cannon and upon the cannonball; however, the velocities imparted upon the cannonball and the cannon will differ because their masses differ. If p1 = p2, then m1v1 = m2v2. If m1 = the 250 kg cannon, and m2 = the 5.0 kg cannonball, then the following observation will be observed:
( m1 / m2 )( v1 ) = v2. Therefore, ( 250 kg / 5.0 kg )( v1 ) = v2. This leaves ( 50 )( v1 ) = v2. The recoil velocity of the cannon ( v1 ) might be small, but the velocity of the cannonball ( v2 ) will be 50 times greater, because the inertia of an object is determined by how much mass it possesses.
Now, let’s examine how angular momentum is similar in principle to the above example. It can be shown that an angle ( θ ) is described by dividing a length ( l ) along the circumference of a circle by the radius ( r ) of the circle. Therefore, ( l / r ) = ( θ ). If a length ( l ) along the circumference of a circle is equal to the circle’s radius, the corresponding angle is referred to as being 1 radian ( 1 rad ). Since the complete distance around a circle = 2𝜋r, the number of radians in a circle = ( 2𝜋r / r ) = 2𝜋, which is close to 6.28 times the length of the circle’s radius ( r ).
For convenience, the angle of a full rotation through a circle is divided into 360 degrees; therefore, 360 degrees = 2𝜋 radians. At an angle of 180 degrees, 𝜋 radians have been traced by the radius along the circle’s circumference. This relationship is extremely useful when converting degrees to radians, and vice-versa. In summary, a radian is some length along the diameter of a circle, and the angle between the beginning and end of that length is defined by some angle ( θ ).
If ( l / r ) = θ, then l = ( r )( θ ). Since distance divided by time = velocity ( v ), we can divide both sides of l = ( r )( θ ) by seconds ( s ) to obtain an expression for angular velocity ( v ); therefore, v = ( r )( ⍵ ), where ⍵ ( omega ) is the angular speed ( θ / s ).
We must now convert the above equation to the familiar equation for centripetal acceleration ( ac ) = ( v2 / r ):
v = ( r )( ⍵ )
v2 = ( r2 )( ⍵2 )
( v2 / r ) = ( ac ) = ( r )( ⍵2 )
Since F = ( m )( a ), the centripetal ( center seeking ) force ( Fc ) = ( m )( r )( ⍵2 ).
Work = ( force x distance ), and the quantity of energy needed to perform work is measured in joules ( J ). Within rotating systems, rotational motion is maximized when an applied force is perpendicular to a radius that extends from a central axis of rotation; thus, the energy of a spinning system = ( Fc )( r sin θ ) = ( m )( r2 )( ⍵2 ) when the angle ( θ ) between the applied force and the radius is 900. This is the equation that describes torque ( 𝝉 ) of a system. The ( m )( r2 ) term is the ” moment of inertia ” ( I ) of a spinning system. The moment of inertia ( I ) defines a spinning object’s ability to resist a change in motion. For this reason, torque can be expressed as ( 𝝉 ) = ( I )( ⍵2 ). A SPINNING FIGURE SKATER WITH OUTSTRETCHED ARMS WILL SPIN MORE RAPIDLY IF THEY PULL THEIR ARMS INWARD, because they effectively decrease their moment of inertia ( I ) upon doing so. Momentum must be conserved, and a loss of inertia within a spinning system is offset by an increase in angular speed ( ⍵ ).
The average kinetic energy of a rotating system is KEavg = ( ½ I )( ⍵2 ). Part b. of the question asks for the before and after kinetic energy ( KE ) of the system. In order to solve part a. of the question, we need a formula for angular momentum ( L ). Since angular momentum is defined relative to some distance ( r ) from an axis of rotation, L = ( p )( r ) = ( m )( v )( r ). It is easier to apply torque to a wrench that has a long handle as opposed to a shorter one. Therefore, if angular velocity ( v ) = ( r )( ⍵ ), then L = ( m )( r⍵ )( r ) = ( m )( r2 )( ⍵ ). Since the ( m )( r2 ) term is the moment of inertia ( I ), L = ( I )( ⍵ ).
In this problem, the student initially holds two 3.09 kg masses at a distance ( ri = 0.99 m ) from the axis of rotation. The moment of inertia of the system ( If ) will differ from ( Ii ) as a consequence of the dumbbells being pulled inward a distance ( rf = 0.298 m ) from the axis of rotation. We must first obtain the SUM of the initial moments of inertia for all rotating components of the system ( student, stool, and dumbbells ); therefore, Ii = Ii1 + Ii2 + Ii3 = [ ( m1 )( r12 ) + ( m2 )( r22 ) + ( m3 )( r33 ) ].
Let’s allow Ii1 to equal the moment of inertia of the student + stool = ( 2.52 kg*m2 ). Now, Ii2 + Ii3 = ( mi2 )( ri22 ) + ( mi3 )( ri32 ). Since mi2 = mi3, we factor mass out of the summation: ( mi )( ri22 + ri32 ). Furthermore, since ri2 = ri3, ( mi )( ri2 + ri2 ) = ( mi )( 2ri2 ) = Ii. The sum of Ii1 + Ii2 + Ii3 = [ ( 2.52 kg*m2 ) + ( mi )( 2ri2 ) ]. Inserting the dumbbell mass into the equation yields Ii = [ ( 2.52 kg*m2 ) + ( 3.09 kg )( 2 )( 0.99m2 ) ], so Ii = [ ( 2.52 kg*m2 ) + ( 6.06 kg*m2 ) ] = 8.58 kg*m2. The system rotates with an initial angular speed of ( 0.749 rad / s ) = ⍵i. Therefore, the initial angular momentum of the system Li = ( Ii )( ⍵i ) = ( 8.58 kg*m2 )( 0.749 rad / s ) = ( 6.43 kg*m2*s-1 ).
We now find the final angular speed ( ⍵f ) of the system by setting Li = Lf :
( 6.43 kg*m2*s-1 ) = ( If )( ⍵f )
Furthermore,
If = If1 + If2 + If3
Since the inertia of the student + chair did not change, If1 = Ii1 = ( 2.52 kg*m2 ), and If2 + If3 = [ ( mf2 )( rf22 ) + ( mf3 )( rf32 ) ]. The dumbbell masses did not change, their masses may once again be factored out of If2 + If3: [ ( m )( rf22 ) + ( rf32 ) ]. The final radial distance for both dumbbells from the axis of rotation is the same; thus, If2 + If3 = [ ( m )( rf2 + rf2 ) ] = [ ( m )( 2rf2 ) ]. Substitution of previously derived values gives us If = If1 + If2 + If3 = [ ( 2.52 kg*m2 ) + ( 3.09 kg )( 2 )( 0.298m2 ) ]. Therefore, If = If1 + If2 + If3 = ( 2.52 kg*m2 + 0.54 kg*m2 ) = 3.06 kg*m2.
Recall that Ii = 8.58 kg*m2. The smaller value of If = 3.06 kg*m2 shows that the final system has less inertia ( resistance to a change in ⍵ ); thus, the final angular speed ( ⍵f ) should be greater than the initial angular speed ( ⍵i = 0.749 rad / s ):
Li = Lf
Li = ( If )( ⍵f )
( 6.43 kg*m2*s-1 ) = ( 3.06 kg*m2 )( ⍵f )
[ ( 6.43 kg*m2*s-1 ) / ( 3.06 kg*m2 ) ] = ( ⍵f ) = ( 2.10 rad / s ).
ANSWER ( a ) : ⍵f = 2.10 rad / s.
THE INCREASE IN ANGULAR SPEED FROM ( 0.749 rad / s ) TO ( 2.10 rad / s ) IS CONSISTENT WITH A DECREASE IN THE SYSTEM’S MOMENT OF INERTIA.
We now have values for the initial and final angular speeds of the system:
( 0.749 rad / s ) and ( 2.10 rad / s )
The kinetic energy ( KE ) of a revolving system is ( ½ I )( ⍵2 ). Since ( Ii ) = ( 8.58 kg*m2 ), the initial kinetic energy of the system ( KEi ) = ( ½ 8.58 kg*m2 )( 0.749 rad / s )2 = ( 2.41 J ).
ANSWER b1: KEi = 2.41 J.
The final kinetic energy ( KEf ) = ( ½ 3.06 kg*m2 )( ( 2.10 rad / s )2 = ( 6.75 J ).
ANSWER b2: KEf = 6.75 J.
A DECREASE IN THE SYSTEM’S MOMENT OF INERTIA LED TO AN INCREASE IN KINETIC ENERGY OF THE SYSTEM.
RENAISSANCE PHYSICS 