Answer:
F2 = 834 N
Explanation:
We are given the following for the bicycle;
Diameter; d1 = 63 cm = 0.63 m
Mass; m = 1.75 kg
Resistive force; F1 = 121 N
For the sprocket, we are given;
Diameter; d2 = 8.96 cm = 0.0896 m
Radius; r2 = 0.0896/2 = 0.0448 m
Radial acceleration; α = 4.4 rad/s²
Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²
Where r1 = (d1)/2 = 0.63/2
r1 = 0.315 m
Thus, I = 1.75 × 0.315²
I = 0.1736 Kg.m²
The torque is given by the relation;
I•α = F1•r1 - F2•r2
Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².
Thus;
0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)
>> 0.76384 = 38.115 - (0.0448F2)
>> 0.0448F2 = 38.115 - 0.76384
>> F2 = (38.115 - 0.76384)/0.0448
>> F2 = 833.73 N
Approximately; F2 = 834 N
