Interpreting force in rotational motion

[MathewsMD]

56. A motor is connected to a solid cylindrical drum with diameter 1.2 m and mass 51 kg. A massless rope is attached to the drum and tied at the other end to a 38-kg weight, so the rope will wind onto the drum as it turns. What torque must the motor apply if the weight is to be lifted with acceleration 1.1 m/s²?

Solution:

I understand almost all the steps. The part where I am confused is when they say F_app = m(a+g), which makes sense on its own. But then in the next part, they τ_app + τ_mass = 1/2MR(a/R) where τ_app = Rm(a+g). Why is this not just RF_g which is equal to Rmg? They way they answered the question makes me think they already took into account the torque applied into the torque due to mass since its linear acceleration, by itself, is only g but they included a which is from the applied torque.

Any help is greatly appreciated!

 

[lucasem_]

You know that [itex]a[/itex] in [itex]F=ma=F_{app}-mg[/itex] is [itex]a=1.1m/s^2[/itex]. Because [itex]F_{app}[/itex] is known [itex]\left(\; F_{app}=m(a+g)\;\right)[/itex], and is applied with moment arm [itex]R[/itex], the torque for the entire mass is [itex]\tau_{mass}=-RF_{app}= -Rm(a+g)[/itex]. Because the net torque [itex]\tau_{net}=I\alpha[/itex] where [itex]\alpha=\frac{a}{R}[/itex] (because it's just angular acceleration and you know that a distance [itex]R[/itex] from the center, the linear acceleration is [itex]a[/itex]), you could see now how [tex]\tau_{net}=\tau_{app}+\tau_{mass}\\
\Rightarrow \tau_{app}=\tau_{net}-\tau_{mass}= I \frac{a}{R} -\left(-Rm(a+g)\right)[/tex]

edit: You account for [itex]F_g[/itex] in [itex]F_{app}[/itex], because the block isn't just falling. There is another force applied so that it's net acceleration is [itex]a[/itex]. This is why they don't just use [itex]RF_g[/itex], but they use [itex]RF_{app} = Rma - RF_g[/itex]

 

[MathewsMD]

You know that [itex]a[/itex] in [itex]F=ma=F_{app}-mg[/itex] is [itex]a=1.1m/s^2[/itex]. Because [itex]F_{app}[/itex] is known [itex]\left(\; F_{app}=m(a+g)\;\right)[/itex], and is applied with moment arm [itex]R[/itex], the torque for the entire mass is [itex]\tau_{mass}=-RF_{app}= -Rm(a+g)[/itex]. Because the net torque [itex]\tau_{net}=I\alpha[/itex] where [itex]\alpha=\frac{a}{R}[/itex] (because it's just angular acceleration and you know that a distance [itex]R[/itex] from the center, the linear acceleration is [itex]a[/itex]), you could see now how [tex]\tau_{net}=\tau_{app}+\tau_{mass}\\
\Rightarrow \tau_{app}=\tau_{net}-\tau_{mass}= I \frac{a}{R} -\left(-Rm(a+g)\right)[/tex]

edit: You account for [itex]F_g[/itex] in [itex]F_{app}[/itex], because the block isn't just falling. There is another force applied so that it's net acceleration is [itex]a[/itex]. This is why they don't just use [itex]RF_g[/itex], but they use [itex]RF_{app} = Rma - RF_g[/itex]

Thanks for the clarification!