Question
An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2.4 kg and radius 23.02 cm. The right wheel has a mass of 2.8 kg and radius 32.21 cm. The hanging mass on the left is 1.97 kg and on the right 1.35 kg. What is the acceleration of the hanging mass on the left? Use down as the positive direction. The acceleration of gravity is 9.8 m/s^2. Answer in units of m/s^2.
Answers
GPT-4o mini
To find the acceleration of the hanging mass on the left in an Atwood machine with differing wheel characteristics, we need to analyze the forces acting on both masses and account for the rotational motion of the wheels.
### Information Given:
- Mass of left wheel (\(M_L\)): 2.4 kg
- Radius of left wheel (\(R_L\)): 0.2302 m (23.02 cm)
- Mass of right wheel (\(M_R\)): 2.8 kg
- Radius of right wheel (\(R_R\)): 0.3221 m (32.21 cm)
- Mass hanging on the left (\(m_L\)): 1.97 kg
- Mass hanging on the right (\(m_R\)): 1.35 kg
- Acceleration due to gravity (\(g\)): 9.8 m/s²
### Step 1: Calculate moments of inertia for both wheels
The moment of inertia \(I\) of a wheel (cylinder) with mass \(M\) and radius \(R\) is given by:
\[
I = MR^2
\]
Calculating for the left wheel:
\[
I_L = M_L R_L^2 = 2.4 \, \text{kg} \times (0.2302 \, \text{m})^2 = 2.4 \times 0.0525 \approx 0.1260 \, \text{kg m}^2
\]
Calculating for the right wheel:
\[
I_R = M_R R_R^2 = 2.8 \, \text{kg} \times (0.3221 \, \text{m})^2 = 2.8 \times 0.1039 \approx 0.2910 \, \text{kg m}^2
\]
### Step 2: Establish equations of motion
Let \(a\) be the acceleration of the left mass, and let \(\alpha\) be the angular acceleration of the wheels related by:
\[
a = R \alpha
\]
For the left mass (\(m_L\)):
\[
m_L g - T_L = m_L a
\]
(1) where \(T_L\) is the tension in the left side.
For the right mass (\(m_R\)):
\[
T_R - m_R g = -m_R a
\]
(2) where \(T_R\) is the tension in the right side.
### Step 3: Relate tensions to torques on the wheels
The net torque (\(\tau\)) on each wheel caused by tensions:
\[
\tau_L = T_L R_L
\]
\[
\tau_R = T_R R_R
\]
In terms of angular acceleration, we have:
\[
T_L R_L = I_L \alpha \quad \Rightarrow \quad T_L = \frac{I_L \alpha}{R_L}
\]
\[
T_R R_R = I_R \alpha \quad \Rightarrow \quad T_R = \frac{I_R \alpha}{R_R}
\]
### Step 4: Solve the system of equations
Substituting \(a = R_L \alpha\) and \(a = -R_R \alpha\) gives:
From (1):
\[
m_L g - \frac{I_L a}{R_L^2} = m_L a
\]
From (2):
\[
\frac{I_R a}{R_R^2} - m_R g = -m_R a
\]
Combining these, we will express both equations in terms of \(a\):
\[
m_L g - m_L a = \frac{I_L a}{R_L^2}
\]
\[
\frac{I_R a}{R_R^2} = m_R g + m_R a
\]
Now, these can be simplified and put together in terms of \(a\):
Combine the equations:
\[
m_L g + m_R g = m_L a + m_R a + \frac{I_L a}{R_L^2} + \frac{I_R a}{R_R^2}
\]
Substituting the known values of the moments of inertia calculated earlier, we find:
\[
m_L g + m_R g = (m_L + m_R + \frac{I_L}{R_L^2} + \frac{I_R}{R_R^2})a
\]
### Step 5: Plug in values to find \(a\)
Calculating the total masses:
\[
(1.97 + 1.35) \times 9.8 = (3.32 + \frac{0.1260}{(0.2302)^2} + \frac{0.2910}{(0.3221)^2})a
\]
Calculate the inertia terms:
\[
\frac{0.1260}{0.0530} \approx 2.38 \quad \text{and} \quad \frac{0.2910}{0.1035} \approx 2.81
\]
Thus:
\[
(3.32) \times 9.8 = (3.32 + 2.38 + 2.81)a
\]
From this:
\[
32.556 = (8.51) a \quad \Rightarrow \quad a = \frac{32.556}{8.51}
\]
Calculating the value of \(a\):
\[
a \approx 3.83 \, \text{m/s}^2
\]
The final answer is:
\[
\boxed{3.83 \, \text{m/s}^2}
\]
### Information Given:
- Mass of left wheel (\(M_L\)): 2.4 kg
- Radius of left wheel (\(R_L\)): 0.2302 m (23.02 cm)
- Mass of right wheel (\(M_R\)): 2.8 kg
- Radius of right wheel (\(R_R\)): 0.3221 m (32.21 cm)
- Mass hanging on the left (\(m_L\)): 1.97 kg
- Mass hanging on the right (\(m_R\)): 1.35 kg
- Acceleration due to gravity (\(g\)): 9.8 m/s²
### Step 1: Calculate moments of inertia for both wheels
The moment of inertia \(I\) of a wheel (cylinder) with mass \(M\) and radius \(R\) is given by:
\[
I = MR^2
\]
Calculating for the left wheel:
\[
I_L = M_L R_L^2 = 2.4 \, \text{kg} \times (0.2302 \, \text{m})^2 = 2.4 \times 0.0525 \approx 0.1260 \, \text{kg m}^2
\]
Calculating for the right wheel:
\[
I_R = M_R R_R^2 = 2.8 \, \text{kg} \times (0.3221 \, \text{m})^2 = 2.8 \times 0.1039 \approx 0.2910 \, \text{kg m}^2
\]
### Step 2: Establish equations of motion
Let \(a\) be the acceleration of the left mass, and let \(\alpha\) be the angular acceleration of the wheels related by:
\[
a = R \alpha
\]
For the left mass (\(m_L\)):
\[
m_L g - T_L = m_L a
\]
(1) where \(T_L\) is the tension in the left side.
For the right mass (\(m_R\)):
\[
T_R - m_R g = -m_R a
\]
(2) where \(T_R\) is the tension in the right side.
### Step 3: Relate tensions to torques on the wheels
The net torque (\(\tau\)) on each wheel caused by tensions:
\[
\tau_L = T_L R_L
\]
\[
\tau_R = T_R R_R
\]
In terms of angular acceleration, we have:
\[
T_L R_L = I_L \alpha \quad \Rightarrow \quad T_L = \frac{I_L \alpha}{R_L}
\]
\[
T_R R_R = I_R \alpha \quad \Rightarrow \quad T_R = \frac{I_R \alpha}{R_R}
\]
### Step 4: Solve the system of equations
Substituting \(a = R_L \alpha\) and \(a = -R_R \alpha\) gives:
From (1):
\[
m_L g - \frac{I_L a}{R_L^2} = m_L a
\]
From (2):
\[
\frac{I_R a}{R_R^2} - m_R g = -m_R a
\]
Combining these, we will express both equations in terms of \(a\):
\[
m_L g - m_L a = \frac{I_L a}{R_L^2}
\]
\[
\frac{I_R a}{R_R^2} = m_R g + m_R a
\]
Now, these can be simplified and put together in terms of \(a\):
Combine the equations:
\[
m_L g + m_R g = m_L a + m_R a + \frac{I_L a}{R_L^2} + \frac{I_R a}{R_R^2}
\]
Substituting the known values of the moments of inertia calculated earlier, we find:
\[
m_L g + m_R g = (m_L + m_R + \frac{I_L}{R_L^2} + \frac{I_R}{R_R^2})a
\]
### Step 5: Plug in values to find \(a\)
Calculating the total masses:
\[
(1.97 + 1.35) \times 9.8 = (3.32 + \frac{0.1260}{(0.2302)^2} + \frac{0.2910}{(0.3221)^2})a
\]
Calculate the inertia terms:
\[
\frac{0.1260}{0.0530} \approx 2.38 \quad \text{and} \quad \frac{0.2910}{0.1035} \approx 2.81
\]
Thus:
\[
(3.32) \times 9.8 = (3.32 + 2.38 + 2.81)a
\]
From this:
\[
32.556 = (8.51) a \quad \Rightarrow \quad a = \frac{32.556}{8.51}
\]
Calculating the value of \(a\):
\[
a \approx 3.83 \, \text{m/s}^2
\]
The final answer is:
\[
\boxed{3.83 \, \text{m/s}^2}
\]