Question

An Atwood machine is constructed using a disk of mass 2 kg and radius 24.8 cm. What is the acceleration of the system? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .

Answer 1

Answer:

a = (m₂-m₁) / (m₂ + m₁ + ½ m)

a = (m₂-m₁) / (m₂ + m₁ + 1)

Explanation:

An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.

Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive

Equation of the side of m₁

              T₁ - W₁ = m₁ a

Equation of the side of m₂

              W₂ - T₁ = m₂ a

Mass pulley equation m; by convention the counterclockwise rotation is positive

              τ = I α

              T₁ R - T₂ R = I (-α)

The moment of inertia of a disk is

              I = ½ m R²

Angular and linear acceleration are related

             a = α R

             α = a / R

The rotation is clockwise, so it is negative

We replace

             (T₁ –T₂) R = ½ m R² (-a / R)

             T₁ -T₂ = - ½ m a

Let's write our three equations together

              T₁ - m₁ g = m₁ a

              m₂ g - T₂ = m₂ a

              T₁ –T₂ = -½ m a

Let's multiply the last equation by (-1) and add

               m₂ g - m₁ g = m₂ a + m₁ a + ½ m a

                a = (m₂-m₁) / (m₂ + m₁ + ½ m)

calculate

               a = (m₂ - m₁)/ (m₁ +m₂ + 1)

Answer 2

Based on the calculations, the acceleration of this system is equal to 0.58 $m/s^2$.

Given the following data:

• Mass of disk = 2 kg.

• Radius of disk = 24.8 cm.

• Acceleration of gravity = 9.8 $m/s^2$

How to calculate the acceleration of the system.

First of all, we would determine the moment of inertia of this disk by using this formula:

$I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2$

Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:

For the first force:

$F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a$

For the second force:

$F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a$

For the torque, we have:

$\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}$

Acceleration, a = 0.58 $m/s^2$

Read more on acceleration here: brainly.com/question/24728358