Question

A bicycle wheel has a diameter of 63.9 cm and a mass of 1.86 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 123 N is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a 8.96 cm diameter sprocket if the wheel is to attain an acceleration of 4.42 rad/s2? (b) What force is required if the chain shifts to a 5.50 cm diameter sprocket?

Answer 1

here,

mass of the wheel , m = 1.86 kg

diameter of the wheel , d = 63.9 cm = 0.639 m

radius , r = 0.3195 m

force , F = 123 N

(a)
  
Torque = Inertia * alpha (angular acceleration rad/s^2)

Torque also equals force * radius

inertia of hoop = mr^2 = 1.86 * 0.3195^2 = 0.18987 kg-m^2

Torque = 0.18987 * 4.42 = 0.8392 Nm

diameter of the sporcket , d' = 0.0896 m

r' = 0.0448 m

0.8392/r' = force

force = 0.8392/0.0448 = 18.73 N

for the resistive force

f * radius of the rim = radius of the sprocket * f'

123 * 0.3195 = f' * 0.0448

f' = 877.2 N

Total = f' + force = 877.2 + 18.73 = 895.93 N

the applied force must be 895.93 N

b)

  
Torque = Inertia * alpha (angular acceleration rad/s^2)

Torque also equals force * radius

inertia of hoop = mr^2 = 1.86 * 0.3195^2 = 0.18987 kg-m^2

Torque = 0.18987 * 4.42 = 0.8392 Nm

diameter of the sporcket , d' = 0.055 m

r' = 0.0275 m

0.8392/r' = force

force = 0.8392/0.0275 = 30.52 N

for the resistive force

f * radius of the rim = radius of the sprocket * f'

123 * 0.3195 = f' * 0.0275

f' = 1427.69 N

Total = f' + force = 1427.69 + 30.52 = 1458.21 N

the applied force must be 1458.21 N