Question

1. When cycling Dr. Leary can continuously exert 159 ft-lbs of torque on the front sprocket (with 53 teeth) while pedaling at a speed of 126-rpm. What is the increased speed and decreased torque when the selected rear sprocket has 42 teeth, and what is the further increased speed and further decreased torque when the selected rear sprocket has 18 teeth?

Answer 1

Solution:

Torque on front sprocket = 159 ft-lbs

No of teeths in front sprocket = 53

Pedaling speed = 126 rpm

If selected rear s procket has 42 teeths we have to find out the change in torque and speed.

Now, the gear ratio can be defined as the ratio of number of teeths of front sprocket to the number of teeths of rear s procket

Thus, Gear ratio = 53/42 = 1.26

Now, the change in rotational speed = pedaling speed X Gear ratio

Or, change in speed = 126 rpm X 1.26 = 159 rpm

Thus the speed will be increased.

Now as per the law of gears, rotational speed of gears is inversely proportional to the torque

Thus,

V1/V2 = T2/T1

Where,

V1 = initial rotational speed

V2 = final rotational speed

T2 = Final torque

T1 = initial torque

Thus, change in torque (T2) = V1/V2 * T1

Or, T2 = 126/159 * 159 ft-lbs

Or, T2 = 126 ft-lbs

Similarly, if number of teeths of rear sprocket is changed to 18 teeths

Gear ratio = 53/18 = 2.94

Thus, increase in speed = 2.94 x 126 rpm = 371 rpm

Decrease in torque = 126/371 * 159 ft-lbs = 54 ft-lbs