Rolling motion: Bicycle
different cases for the wheel

1. Wheel at rest

Along a vertical axis, we have:     N + mg = 0

2. Motion of the wheel with sliding effect

The whell slides witout friction as it does a bicycle when it is sliding on ice. This is the case of a simple block that slides along a frictionless path.
All the parts of the whell has the same linear velocity v_cm. The angular velocity ω is zero, so is the angular acceleration.
With some external force, the whell will undergo a linear acceleration a = dv_cm/dt.

The whell slides with friction as it does a bicycle when it is moving and braked all of a sudden, or as it does an erasing rubber on a paper. This is the case of a simple block that slides along a path with kinetic friction.

All the parts of the whell has the same linear velocity v_cm. The angular velocity ω is zero, so is the angular acceleration.
The external force acting on the wheel is the friction force ƒ providing the whell a linear decceleration a = dv_cm/dt.

Along a horizontal axis, we have :
- ƒ + ma = 0 or ƒ = ma

Along a vertical axis, we have :
N = m g
Therefore:
ƒ = μ_k N = μ_k m g

    ƒ = μ_k N = μ_k m g = m a

    a = μ_k g

3. Rotation of a free wheel

The wheel rotates freely about its axis located at the center of the wheel.
Its angular velocity ω is related to a tangential speed v as: v = ω r, where r is the radius of the whell.

The translation motion of the wheel is zero.

Therefore:
a = d/v_cmdt = 0

    a = d/v_cmdt = 0

    v_t = ω r

    α = dω/dt

    a_t = r α

4. Frictionless rotating wheel

The frictionless wheel rotates and slides about its axis located at the center of the wheel. The wheel floats on its support, as on ice.
Its angular velocity ω is related to a tangential speed v as: v = ω r, where r is the radius of the whell.

The rotating motion of the wheel will remain uniforme (ω = constant) unless a torque is exerted on it and provides it an angular acceleration.

    v_t = ω r

    α = dω/dt

    a_t = r α

5. Friction rotating wheel

The wheel rotates with a kinetic friction about its axis located at its center, as it does a rotating sharpening tool when we hold someting against its grindstone.
Its angular velocity ω is related to a tangential speed v as: v = ω r, where r is the radius of the whell.

Its angular acceleration α has the expression:
α = dω/dt.
Related to the torque due to the friction force ƒ , we have:
τ_net = ƒ r =I α ,
where I is the moment of inertia of the wheel.

6. Rolling and slading frictionless wheel

7. Rolling and slading wheel with friction

8. Rolling frictionless wheel

The wheel rolls (translates and rotates) without friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.
And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

9. Rolling wheel with friction

The wheel rolls (translates and rotates) with friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.
Related to the torque due to the friction force ƒ , we have:
τ_net = ƒ r = I α ,
where I is the moment of inertia of the wheel.


And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = ƒ r = I α

10. Rolling braked frictionless wheel

The wheel rolls (translates and rotates) without friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.
Related to the torque due to the brake force b , we have:
τ_net = - b r = I α ,
where I is the moment of inertia of the wheel.


And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = - b r = I α

11. Rolling braked wheel with friction

The wheel rolls (translates and rotates) with friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.
Related to the two torques due to friction force ƒ and the brake force b , we have:
τ_net = ƒ r - b r = I α ,
where I is the moment of inertia of the wheel.


And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = ƒ r - b r = I α

12. Rolling pulled frictionless wheel

The wheel rolls (translates and rotates) without friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.

And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt,
with
F = m a_cm, m is the mass of the wheel.

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    F = m a_cm

13. Rolling pushed wheel with friction

The wheel rolls (translates and rotates) without friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt. Related to the torque due to the static friction force ƒ, we have:
τ_net = ƒ r = I α ,
where I is the moment of inertia of the wheel.



And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt,
with
F = m a_cm, m is the mass of the wheel.

F is responsible for a_cm, and ƒ is pesponsible for α

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = ƒ r = I &alpha

    F = m a_cm

14. Rolling frictionless wheel with tension on the sprocket

The wheel rolls (translates and rotates) without friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt.
Related to the torque due to the tension force T , we have:
τ_net = T r1 = I α ,
where I is the moment of inertia of the wheel, and r1 is the radius of the sprocket.

And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt,

The tension T is pesponsible for α

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = T r1 = I α

15. Rolling wheel with tension on the sprocket and friction

The wheel rolls (translates and rotates) with friction. It translates along its center of mass which is its center. Plus, it rotates about its axis located at this center.

Its angular velocity ω is related to a tangential speed v_t as:
v_t= ω r,
where r is the radius of the whell.

We have also:
v_t = v_cm

The relative speed v at the top of the wheel v is the linear sum of v_t and v_cm
v = v_t + v_cm = 2 v_cm.

Its angular acceleration α has the expression:
α = dω/dt. Related to the two torques due to the friction force ƒ and the tension on the sprocket provided by a pedalling system, we have:
τ_net = T r - ƒ r = I α ,
where I is the moment of inertia of the wheel.

And its tangential acceleration is equal to its linear acceleration:
a = a_t = a_cm = dv_cm/dt,

The tension T is pesponsible for α, and The friction ƒ , by the torque ƒ r, is pesponsible for α.

    v_t= ω r

    v_t = v_cm

    v = v_t + v_cm = 2 v_cm

    a = a_t = a_cm = dv_cm/dt = α r

    τ_net = T r - ƒ r = I α

We would have T r > ƒ r to move on.

16. Acceleration of a bicycle

We have the following relationships:
F r_p = T r₂    (1)
T r₁ - ƒ_back r = I α    (2)
(F r_p/r₂) r₁ - ƒ_back r = I α    (3)
(F r_p r₁/r₂) - ƒ_back r = I α    (4)
F_net = ƒ_back - ƒ_front = m a = m r    (5)

m is the mass of the bicycle and a is its linear or tangential acceleration.
α is the angular acceleration of the back as well as of the front wheel. We assume that the two wheels have approximately the same moment of inertia I.

ƒ_front r = I α    (6)

Therefore:
(5) becomes:
ƒ_back r - I α = m a = m r² α(5)
With (3)
(F r_p/r₂) r₁ - 2I α = m r² α(5)
(F r_pr₁/r₂) = 2I α + m r² α(5)
(F r_pr₁/r₂) = (2I + m r²) α(5)
α = (F r_pr₁/r₂) /(2I + m r²)
a = ( r F r_pr₁/r₂) /(2I + m r²)

    a = ( r F r_pr₁/r₂) /(2I + m r²)