Question

Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown such that the angular acceleration of the two wheels is the same.
1)Compare the magnitudes of the two forces
F2 = F1
F2 = 2 F1
F2 = 4 F1

Answer 1

Answer:

The force on second wheel is twice off the force on first wheel.

Explanation:

In this case, two wheels can rotate freely about fixed axles through their centers. We know that, in rotational mechanics, the torque is given by :

$\tau=I\alpha$

Also, $\tau=Fr$

And moment of inertia is, $I=mr^2$

It implies,

$Fr=mr^2\alpha \\\\F=mr\alpha$    

Here, one has twice the radius of the other. Ratio of forces will be :

$\dfrac{F_1}{F_2}=\dfrac{mr_1\alpha }{mr_2\alpha }\\\\\dfrac{F_1}{F_2}=\dfrac{r_1 }{2r_1}\\\\\dfrac{F_1}{F_2}=\dfrac{1 }{2}\\\\F_2=2F_1$

So, the force on second wheel is twice off the force on first wheel.