ROTATIONAL MOTION: At What Rate will the Yo-Yo Accelerate?

Several forces must be taken into account to study the motion of a yo-yo. If we assume a hand to be stationary when a yo-yo begins its descent, a tension force acts upward upon the yo-yo’s string. Opposite to the tension force is the force exerted upon the system by the gravitational force of attraction ( F_w = mg ). These forces contribute to the torque exerted on the yo-yo’s center of mass:

In systems where a = 0, the sum of all forces acting upon the system is zero. Since the system above moves downward with linear acceleration ( a ), a net force acts upon the system:

∑ F = ma

Notice that the system has both linear acceleration ( a ) and angular acceleration ( 𝞪 ) components. The tension force ( T = F_t ) exerts a torque ( 𝜏 ) on the yo-yo’s central axis. Although torque is commonly referred to as a “ twisting force “, its units are ( N*m ). The general formula for torque is ( Fr sin θ ). There is a 90⁰ angle between the yo-yo’s string and the central axis of rotation. Since sin 90⁰ = 1, the torque equation may be simplified:

 𝜏 = Fr

The linear acceleration of the system is directly proportional to its angular acceleration. This relationship is derived from the fact that linear velocity ( v ) is directly proportional to angular speed ( ⍵ ), and a = Δv/Δt:

v = r⍵

a = v/t = ( r )( ⍵/t ) = r𝞪

a = r𝞪

F = ma = mr𝞪

mr𝞪

𝜏 = ( mr𝞪 )( r ) = mr² 𝞪

𝜏 = I𝞪

The moment of inertia ( I = mr² ) is a measure of a system’s ability to resist changes in rotational motion. The value of ( I ) differs for objects with different shapes. The moment of inertia and torque values for a yo-yo may be approximated as follows:

I = ½ mr²

𝜏 = ( ½ mr² )( 𝞪 )

Care must be taken when choosing the appropriate value for ( r ); ( I ) represents the entire system’s ability to resist a change in angular motion. The radius that traverses the entire system is ( r_out ), and it is the appropriate value used to determine the system’s moment-of-inertia: 

I = ½ mr_out²

𝜏 = ( ½ mr_out² )( 𝞪 )

The right-hand-side of the equation describes the applied torque’s effect on the system. Notice, however, that the torque that is applied to the system is applied by ( F_t ) acting perpendicular to ( r_in ):

 𝜏 = F_tr_in

F_tr_in = ( ½ mr_out² )( 𝞪 )

In order to solve for the translational acceleration of the system, angular acceleration must be rewritten in terms of its linear acceleration counterpart and substituted for ( 𝞪 ):

a = r𝞪

a/r = 𝞪

Once again, ( r_in ) is the radius that has a force applied to it, and its value is substituted appropriately:

F_tr_in = ( ½ mr_out² )( a/r_in )

F_t = ( ½ mr_out² )( a/r_in²)

F_t = ( ½ m )( r_out/r_in)²( a )

We are now ready to evaluate how the gravitational force of acceleration ( F_w ) contributes to the motion of the system. We are at liberty to assign a positive or negative value to the upward or downward direction. The system accelerates linearly in the same direction as the force of gravity acting upon it. As a consequence, if the gravitational constant of acceleration ( g ) is assigned a positive value, the value of the system’s acceleration ( a ) must be positive as well:

F_w = mg

F = ma

The only force opposing the downward acceleration of the system is the tension force ( T = F_t ). Thus, ( F_t ) is assigned a negative value in the derivation for the net translational force acting upon the system:

mg – F_t = ma

F_t = mg – ma

F_t = ( m )( g – a )

We now have two solved equations for ( F_t ), and we may set them equal to one another:

( m )( g – a ) = ( ½ m )( r_out/r_in )²( a )

Recall that ( r_in = ⅓ r_out ), and thus, ( 3 = r_out/r_in ). Substitution of ( 3 ) into the equation above yields the following results:

( m )( g – a ) = ( ½ m )( 3 )²( a )

( m )( g – a ) = ( ⁹/₂ m )( a )

( g – a ) = ( ⁹/₂ a )

g = ( ⁹/₂ a + ²/₂ a )

g = ¹¹/₂ a

a = ²/₁₁ g