How can I solve this physics problem?

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 3.09 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.99 m from the axis of rotation and the student rotates with an angular speed of 0.749 rad/s. The moment of inertia of the student plus stool is 2.52 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.298 m from the rotation axis (Figure b).

(a) Find the new angular speed of the student.

 _____ rad/s

 

(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.

Kbefore ____ J

Kafter ____ J

Answer 1

Actually much simpler than previous answer: I_o = initial I_total = 2.52 + 2[(3.09)(.99)² = 8.58 kg•m² ;

initial angular velocity w_° = 0.749 rad ⁄s

I =Final I_total = 2.52 + 2[(3.09)(0.298)² = 3.07 kg•m^{2 ;}

Conservation of Angular momentum - I_oω_o= Iω; solve for w [final angular velocity]; ω = I _oω_o ⁄ I

ω = (8.58)(0.749) / 3.07 = 2.09 rad/s

KE before = .5 I_o ω_o² = .5(8.58)(0.749)² = 2.41 J

KE after = .5 I ω² = .5(3.07)(2.09)² = 6.71 J

Answer 2

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.