Rotational dynamics: Atwood's machine

1.Atwood's machine

Atwood's machine consists of a frictionless and not stretching pulley that hangs two masses attached to a string that does not slip on the pulley, as shown in the figure.
This machine uses Newton’s second law for translational (F = m a) and rotational (Ï„ = I α) motions .

If m is the mass of the pulley of radius r, its moment of inertia is I.
The masses m1 and m2 are pulled by the tensions T1 and T2 respectively.

1.1. Second Newton's law for rotation

The second Newton's law for rotation for the pulley is:
τ_net = Στ = I α    (1)
That is :

T1 r - T2 r = r(T1 - T2) = I α    (1')

α is the angular acceleration of the pulley.

1.2. Second Newton's law for translation

The second Newton's law for translation of the two masses is:

For m1:
m1 g - T1 = m1 a    (4)

For m2:
- m2 g + T2 = m2 a    (5)

From (4):
T1 = m1 g - m1 a = m1 (g - a)    (4')

From (5):
T2 = m2 g + m2 a = m2(g + a)    (5')

1.3. Solving for the tensions and accelerations

The tangential acceleration "a1" for the mass m1 is
a1 = α r    (2)
And
The tangential acceleration "a2" for the mass m2 is
a2 = α r    (3)

Therefore
a1 = a2

a1 = a2 = a

Multiplying (4') by r1 and (5') by r2, and using the formula (1') yields:
r(T1 - T2) = r m1 (g - a) - r m2(g + a) = I α

Using the relationships (2) and (3), we get:
r(T1 - T2) = r m1 (g - α r) - r m2(g + α r) = I α

Developing:
r(T1 - T2) = m1 g r - α m1 r² - m2 g r - α m2 r²

Rearranging to solve for α leads to:
r m1 g - r m2 g - α (m1 + m2) r² = I α
r m1 g - r m2 g = α ((m1 + m2) r² + I)

Therefore:
α = r g (m1 - m2)/((m1 + m2) r² + I)

We have then:

a = α r = g (m1 - m2)/((m1 + m2) + I/r²)

T1 = m₁(g - a)

T2 = m₂(g + a)

1.4. Particular cases:

1.4.1. At the static equilibrium:

At the static equilibrium (for rotation and translation), the system is at rest, then its linear and angular acceleration is zero.
Also, when the system has a constant angular speed, its linear and angular acceleration is zero: α = dω/dt = 0.

Therefore, the formula (1') becomes:
r(T1 - T2) = I 0 = 0, then T1 = T2 .

The two torques Ï„1 = T1 r and Ï„2 = T2 r are equal when the system is at rest or at constant speed.

At the static equilibrium:

a = α r = 0

T1 = T2

1.4.2. The pulley is massless:

In the case of the pulley is massless I = 0.
Therefore:
And T1 = T2 = T, Then:
T = m₁(g - a), and
T = m₂(g + a)
Hence:
T = m₁(g - a) = m₁g[1 - (m1 - m2)/(m1 + m2)] =
2m₁m₂/(m₁ + m₂)

Massless pulley:

I = 0

a = α r = g (m1 - m2)/(m1 + m2)

T1 = T2 = T = 2m₁m₂/(m₁ + m₂)

2. Half Atwood’s machine

It is obtained by eliminating one of the hanging masses and wrap the free end of the string around the pulley.



It is particular case of the Atwood's machine described above, where m2 = 0, T2 = 0, m1 = m and T1 = T.

τ = r T = I α

We have then:
a = α r = r² g m/(m r² + I)
T = m(g - a)

a = α r = r² g m/(m r² + I) =
g /[1 + (I/m r²)]


T = m(g - a)

3. Falling motion of the pulley

Now, the pulley of mass m is falling while the string unrolls off a pulley. The results found above still hold:

a = α r = g /[1 + (I/m r²)]

T = m(g - a)

τ = r T = I α

The moment of inertia "I" corresponds to the kind of a cylinder we use. The maximum moment of inertia that exists is I = m r², that corresponds to a hollow ring; that will reduce a of 50%.

4. Application: A yo-yo

4.1. Dynamics considerations

For the acceleration "a", to overcome the limit of 50% of the acceleration of gravity g, we use a yo-yo with a larger radius R to increase I with the same mass m. The radius r of the cylinder around which the string is wrapped is not changed, and therefore the above results still hold:

a = α r = g /[1 + (I/m r²)]

T = m(g - a)

τ = r T = I α

Let's consider a uniform disk, we have I = (1/2) m R²,
and write : β = R/r.

Therefore:
a = α r = g /[1 + (β²/2)]

β = R/r

a = α r = g /[1 + (β²/2)]


In the case of R r, we will have β = 1, and:
a = α r = (2/3) g , and
T = m(g - a) = m (g - 2g/3) = (1/3) mg

4.2. Energy considerations

Using energy considerations, we can find the linear speed of the center of mass of the yo-yo when it reach a distance h from an initial position moving downward while the string unwinds. The moment of inertia of the cylinder of the yo-yo is I = (1/2) m r².

At the initial position:
PE = m g h
KE = 0

At the final position:
PE = 0
KE = KE (translation) + KE (rotation) =
(1/2) m v_cm² + (1/2) I_cm ω²

As the cylinder of the yo-yo falls without sliding, we have:
v_cm = v_tangential = ω r

Therefore:
KE = (1/2) m v_cm² + (1/2) ((1/2) m r²) (v_cm/r)²
= (1/2) m v_cm² (1 + 1/2) = (3/4) m v_cm²
KE = (3/4) m v_cm²

Hence
m g h = (3/4) m v_cm² . That is:
v_cm = [(4/3) g h]^1/2

v_cm = [(4/3) g h]^1/2

5. Windlass and crate

A crate is tied to a rope which is wrapped around a windlass. The crate has a mass m. The windlass is treated as a cylinder of radius R ans mass M. We exert a force F on the crank. The whole system is used to rise the crate as pulling up a bucket of water from a well.

The torque Ï„_crank = F r we exert on the crank is transmitted to the windlass, where it is exerted a second torque Ï„_crate. Therefore, Newton's second law for rotation for the windlass is:
Σ τ = τ_net = I α =
Ï„_crank - Ï„_crate
I is the moment of inertia of the windlass, and α is its angular acceleration.

Ï„_crate = T R. Hence:
I α = F r - T R

Newton's second law for the translation for the crate is:
T - mg = m a
Where "a" is the linear acceleration of the crate.

That gives with a = α R :
T = m g + m a = mg + m α R

The above relationship I α = F r - T R becomes:
I α = F r - T R = F r - (m g + m α R ) R =
F r - m g R - α m R².

Solving for α yields:
α(I + m R²) = F r - m g R . Hence:
α = (F r - m g R)/(I + m R²)

With I = (1/2) M R², we have:
α = (F r - m g R)/(M/2 + m) R²

α = (F r - m g R)/(I + m R²) =
2(F r - m g R)/(M + 2 m) R²

T = Mg + m a = Mg + m α R

The particular case where there is no force F (no crank), we have with F = 0:

α = - 2(m g)/(M + 2 m)R (clockwise)

a = α R = + 2(m g)/(M + 2 m) (counterclockwise)

T = I α/ R = I a/R² = (1/2) M R² a /R²
= (1/2) M a = (1/2) M 2(m g)/(M + 2 m) =
M m g/(M + 2 m)

Free windlass:

a = α R = + 2 m g/(M + 2 m)

T = M m g/(M + 2 m)