Forces acting on a spool of mass M

[issacnewton]

Homework Statement

Two forces $F_1$ and $F_2$ are applied on a spool of mass $M$, moment of inertia, $I$, and radius $R$ as shown in the figure. If the spool is rolling on the surface, find the ratio of forces, $F_1/F_2$ such that friction between spool and the surface is zero.

Homework Equations

$\tau = I\alpha$ , $\sum F = ma$

The Attempt at a Solution

Assume that the acceleration $a$ is toward right. So Newton's second law tells us that $$F_1 + F_2 = Ma$$ And the net torque on the spool is $$\tau = F_2 \left(\frac{R}{2}\right) - F_1 R = I \alpha $$
The condition for rolling without slipping tells us that $a = \alpha R$. So plugging this in above equations we get, $$ \frac{F_2}{2} - F_1 = \frac{Ia}{R^2} $$ Now let $x = F_1/F_2$, So above equations become $$ xF_2 + F_2 = Ma$$ and $$ \frac{F_2}{2} - xF_2 = \frac{Ia}{R^2} $$ Solving for $x$, we get
$$x = \frac{F_1}{F_2} = \frac{MR^2 - 2I}{2(MR^2 + I)} $$ Does my solution seem right ?

Thanks
$\ddot\smile$

 

[haruspex]

$$ \frac{F_2}{2} - F_1 = \frac{Ia}{R^2} $$

Where did the R² come from? Just R, no?

Edit: Sorry - misread an a as an alpha.

Also, not sure you have been consistent with signs on the accelerations. Which way are you taking as positive for the angular acceleration?

 

[issacnewton]

Since $a = \alpha R$, $\alpha = a/R$ and plugging this into the second equation will get us $R^2$ in the denominator on the right side. Since I assume the spool is accelerating towards right, this is clockwise rotation for the spool, so the angular acceleration will be going inside the page by right hand rule. And linear acceleration of the center of mass of the spool, $a$, is towards right.

 

[haruspex]

Since $a = \alpha R$, $\alpha = a/R$ and plugging this into the second equation will get us $R^2$ in the denominator on the right side. Since I assume the spool is accelerating towards right, this is clockwise rotation for the spool, so the angular acceleration will be going inside the page by right hand rule. And linear acceleration of the center of mass of the spool, $a$, is towards right.

see my edit re R².

For the sign, I am still not sure which rotation you define as positive, clockwise or anticlockwise. Your a=αR implies clockwise, but your torque equation, ½RF₂-RF₁=Iα, implies anticlockwise.

 

[issacnewton]

Yes, you are right. In physics, we take counter clockwise rotations as positive. So since rotation is clockwise here, the magnitude of torque would be $RF_1 - RF_2/2$ and angular acceleration is in the same direction as the net torque, $\tau$. If $\alpha$ is the magnitude of the angular acceleration, then we would have $$ RF_1 - RF_2/2 = I\alpha = \frac{Ia}{R}$$ So if I work it out again, I get
$$x = \frac{F_1}{F_2} = \frac{MR^2 + 2I}{2(MR^2 - I)} $$

Does it look Okay ?

 

[haruspex]

Yes, you are right. In physics, we take counter clockwise rotations as positive. So since rotation is clockwise here, the magnitude of torque would be $RF_1 - RF_2/2$ and angular acceleration is in the same direction as the net torque, $\tau$. If $\alpha$ is the magnitude of the angular acceleration, then we would have $$ RF_1 - RF_2/2 = I\alpha = \frac{Ia}{R}$$ So if I work it out again, I get
$$x = \frac{F_1}{F_2} = \frac{MR^2 + 2I}{2(MR^2 - I)} $$

Does it look Okay ?

Yes, that looks right.