Respuesta :
Answer:[tex]25.97 rad/s^2[/tex]
Explanation:
Given
radius of wheel [tex]r=0.28 m[/tex]
mass of wheel [tex]m=8.80 kg[/tex]
Force [tex]F=32 N[/tex]
Moment of Inertia of solid wheel [tex]I=\frac{mr^2}{2}[/tex]
[tex]I=\frac{8.8\times 0.28^2}{2}[/tex]
[tex]I=0.344 kg-m^2[/tex]
Torque is given by
[tex]\tau =F\times r=I\times \alpha [/tex]
[tex]32\times 0.28=0.344\times \alpha [/tex]
[tex]\alpha =25.97 rad/s^2[/tex]
Force on the axle is 32 N since there is no linear acceleration of the system
using Third law F=32 N
Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle
