The drawing shows a bicycle wheel of radius r resting against a small step whose height is h = r/5. A clockwise torque is applied to the axle of the wheel. As the magnitude of torque increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. Let this torque be T. What is the magnitude of the horizontal component of the acceleration of the centre of the wheel when a torque of 2T is applied? Assume that the wheel doesn’t slip at the edge of the step when this torque is applied (ignore the mass of the spokes)

Moderate

Solution

$\sqrt{r^{2} - {(r - h)}^{2}} = \sqrt{r^{2} - {(r - \frac{r}{5})}^{2}} = \frac{3 r}{5}$
[Step 1]: When the wheel gets just lifted, torque, T=mg(3r/5).
[Step 2]: But when a torque of 2 is applied, we can write
[Step 3]: mg $(3 r / 5) = Ι α$ Where $Ι = 2 m r^{2}$
[Step 4]: $α = (\frac{3 g}{10 r})$
The horizontal acceleration is $= (r - h) α = (r - \frac{r}{5}) α = \frac{4}{5} r α$
[Step 5]: $\frac{4}{5} r (\frac{3 g}{10 r}) = \frac{6 g}{25} = \frac{60}{25} = 2.4 m s^{- 2}$

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