AP PHYSICS: Pulleys, Torque, Tension, and the Moment of Inertia

Two spherical bearings of mass ( m₁ ) and ( m₂ ) are attached by belt to a pulley that is situated between them. An electric motor is also attached to the pulley, and it is positioned between these masses. When the motor is turned on, the pulley provides 15.0 N of tension on the length of the belt between it and the ( m₁ ) mass. The tension on the length of belt between the pulley and ( m₂ ) is 0 N. The motor maintains the aforementioned tension for 10.0 seconds, beginning with both bearings being at rest. The dimensions of the bearings in question are as follows:

m₁ = 12.0 kg

r₁ = 0.500 m

m₂ = 18.0 kg

r₂ = 0.150 m

Q: How much work is done by the motor during the period of time that it supplies tension to the system, assuming that the associated pulley is massless? 

A: W = Fd

It is important that we realize that work ( W ) is quantified with the unit SI of energy, which is the joule ( J ). Although torque ( 𝝉 ) is oftentimes described as a “ twisting force “, it’s unit is identical to that of the joule:

𝝉 = F⟂r = Fr sin θ ( N*m )

If we can determine the net torque exerted on the bearings, we will know how much work was done by the motor. Additionally, energy calculations are independent of time ( t ). If we wanted to know how much power ( P ) was used by the motor over the course of 10.0 seconds, time would indeed be a factor, but we’re being asked how much work was performed; not how much work was performed over a given time interval:

𝝉 = Iα

𝝉 = Fr sin θ 

The pulley interacts with the bearings at a 90^o angle, and for this reason, we can drop the sin θ term:

sin 90^o = 1

𝝉 = Fr 

And,

Fr = Iα

The tension force ( Denoted as T below ) exerts a force ( F_T ) on the system in the following manner:

Since the rate at which ( m₂ ) rotates is determined by its attachment to ( m₁ ), the rate of angular acceleration ( α ) of the entire system is established by ( m₁ ); however, the moment of inertia ( I ) for each mass must be determined individually. In addition to this, we must use an expression for ( I ) that is appropriate for a spherically shaped object:

𝝉_net =  𝝉₁ + 𝝉₂

𝝉_net =  I₁α₁+ I₂α₂

α₁ = α₂

𝝉_net = ( I₁ + I₂ )α

We are now ready to solve for the value of ( α ): 

F_Tr₁ = I₁α

α = ( F_Tr₁ / I₁ )

At this point, the total torque exerted on the system may be evaluated:

𝝉_net = ( I₁ + I₂ )( F_Tr₁ / I₁ )

𝝉_net = [ ( I₁ + I₂ ) / I₁ ]( F_Tr₁ )

We must now substitute the expression above with moment of inertia values for a solid sphere: 

I_sphere = ⅖ mr²

---

I_{1 sphere} = ⅖ ( 12 kg )( 0.500 m )²

I_{2 sphere} = ⅖ ( 18 kg )( 0.150 m )²

---

I_{1 sphere} = ⅖ ( 12 kg )( 0.250 m² )

I_{2 sphere} = ⅖ ( 18 kg )( 0.0225 m² )

---

𝝉_net = [ ( I₁ + I₂ ) / I₁ ]( F_Tr₁ )

𝝉_net = [ ( 1.20 kgm² + 0.162 kgm² ) / 1.20 kgm² ][ ( 15.0 N )( 0.500 m ) ]

𝝉_net = ( 2.56 kgm² / 1.20 kgm² )( 7.50 Nm )

𝝉_net = ( 2.13 )( 7.50 Nm )

𝝉_net = 16 Nm