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)

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Infinity Learn · Accepted Answer

$$r2-(r-h)2$$ $$=r2-r-r52=3r5$$[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 $$3$$]: $$mg(3r/5)=$$Ια          Where  Ι$$=2mr2$$[Step $$4$$]:  α$$=3g10rThe$$ horizontal acceleration is $$=r-h$$α$$=r-r5$$α$$=45r$$α[Step $$5$$]:  $$45r3g10r=6g25=6025=2.4ms$$−$$2$$

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