The drwing shows a bicycle wheel of radius r, resting against a small step whose height is `h = r//5`. A clockwise 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 magniitude 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).

Options

A. `4.8 m//s^(2)`
B. `3 m//s^(2)`
C. `2.4 m//s^(2)`
D. `1.5 m//s^(2)`

Answer

C

Explanation

When the wheel gets just lifted `T = mg(3r//5)`. But when a torque of `2T` is applied, we can write:
`mg(3r//5) = l alpha`,
where `l = 2mr^(2)`
The horizontal acceleration is `r alpha (4//5)`.