Question

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive force of 135 N to the tire's rim. The mass of the wheel is 1.90 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 60.0 cm. A chain passes over a sprocket that has a diameter of 8.75 cm. In order for the wheel to have an angular acceleration of 3.70 rad/s2, what force, in Newtons, must be applied to the chain?

Answer 1

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

               $I = MR_A^2$

Substituting  $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$(Radius of wheel)

              $I = 1.9 * 0.33^2$

                $= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

                      $\tau = FR_B - F_rR_A$

Substituting  135 N for $F_r$ (Force acting on sprocket),$\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     $\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

                   $\tau = \alpha I$

Now equating the two equation for torque

                                $F (0.0435) - 135 (0.33) = \alpha I$    

Making F the subject

                     $F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

                  $F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

                       $= 1041.7N$