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1 year ago

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Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radiu

s 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

1 answer:

Murrr4er [49]1 year ago

7 0

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.

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