An atwood machine is constructed using two
wheels (with the masses concentrated at the
rims). the left wheel has a mass of 2.5 kg and
radius 24.03 cm. the right wheel has a mass
of 2.3 kg and radius 31.38 cm. the hanging
mass on the left is 1.64 kg and on the right
1.27 kg
what is the acceleration of the system?
the acceleration of gravity is 9.8 m/s^2
answer in units of m/s^2

The difference in torque and mass applied determines the acceleration of

the system.

The acceleration is approximately 0.4703 m/s²

Reasons:

The mass of the left wheel = 2.5 kg

Radius of the left wheel = 24.03 cm

Mass of the right wheel = 2.3 kg

Radius of the right wheel = 31.38 cm

Mass on the left = 1.64 kg

Mass on the right = 1.27 kg

Acceleration due to gravity, g ≈ 9.8 m/s²

Required:

The acceleration of the system

Solution:

The given acceleration of the

T₃ - m₃·g = m₃·(-a)...(1)

T₄ - m₄·g = m₄·a...(2)

T₃ - T = m₁·a...(3)

T - T₄ = m₂·a...(4)

Add equation (3) to equation (4) gives;

T₃ - T + (T - T₄) = T₃ - T₄ = m₁·a + m₂·a

Subtracting equation (2) from equation (1) gives;

(T₃ - m₃·g) - (T₄ - m₄·g) = m₃·a + m₄·a

T₃ - T₄ = -m₃·a - m₄·a - (m₄·g - m₃·g)

Which gives;

m₁·a + m₂·a = -m₃·a - m₄·a - (m₄·g - m₃·g)

a·(m₁ + m₂) = -a·(m₃ + m₄) - (m₄·g - m₃·g)

-(m₄·g - m₃·g) = (m₃·g - m₄·g) = a·(m₃ + m₄) + a·(m₁ + m₂) = a·(m₃ + m₄ + m₁ + m₂)

The acceleration is a ≈ 0.4703 m/s²

Learn more here:

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0.47 m/s²

Explanation:

Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.

Draw a four free body diagrams, one for each hanging mass and one for each wheel.

For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₃ − m₃g = m₃(-a)

For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₄ − m₄g = m₄a

For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)

T₃ − T = m₁a

For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)

T − T₄ = m₂a

We now have 4 equations and 4 unknowns.  Let's add the third and fourth equations to eliminate T:

(T₃ − T) + (T − T₄) = m₁a + m₂a

T₃ − T₄ = (m₁ + m₂) a

Now let's subtract the second equation from the first:

(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a

T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a

T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a

Setting these two expressions equal:

(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a

(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g

a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)

Plugging in values:

a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)

a = 0.47 m/s²

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