Question

The 25 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it slips as it rolls. Determine the angular acceleration of the wheel and the acceleration of the wheel's center O. The coefficient of kinetic friction between the wheel and the plane is μk = 0.6.

Answer 1

Answer:

Explanation:

radius of gyration of wheel k then

k² = r²/2

r² = 2 k²

r = √2 k

= 1.414 x .3 m

r = .4242 m

Moment of inertia of wheel

= mass x radius of gyration ²

= 25 x .3 x .3

= 2.25 kg m²

Friction force acting on it ( sliding )

= μmg , μ being coefficient of kinetic friction

This friction force will create linear acceleration in forward direction

Acceleration produced

= μg

= .6 x 9.8

= 5.88 m / s ²

This will also rotate the wheel , angular acceleration being

linear acceleration / radius

= 5.88 /.4242

= 13.86 radian / s²