A solid cylindrical pulley with a mass of m=1.50 kg and a radius of r=0.270 m is free to rotate about its axis. An object of mass 0.270 kg is attached to the pulley with a light string (the figure below). Assume the string does not stretch or slip. Calculate the tension in the string. (Express your answer to three significant figures.)

To calculate the tension in the string, we can apply Newton's second law for both the object hanging from the pulley and the pulley itself.

### Given:
- Mass of the pulley, \( m_p = 1.50 \, \text{kg} \)
- Radius of the pulley, \( r = 0.270 \, \text{m} \)
- Mass of the hanging object, \( m = 0.270 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

### Step 1: Analyze the forces on the hanging mass
For the hanging mass \( m \):
\[
F_{\text{net}} = mg - T = ma
\]
where \( T \) is the tension in the string and \( a \) is the linear acceleration of the mass.

### Step 2: Analyze the pulley
The torque \( \tau \) on the pulley due to the tension in the string is given by:
\[
\tau = T \cdot r
\]
The torque is also related to the angular acceleration \( \alpha \) of the pulley:
\[
\tau = I \alpha
\]
where \( I \) is the moment of inertia of the pulley. For a solid cylinder:
\[
I = \frac{1}{2} m_p r^2
\]
The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is:
\[
a = r \alpha \quad \Rightarrow \quad \alpha = \frac{a}{r}
\]

### Step 3: Substitute into the torque equation
Substituting for \( \alpha \):
\[
T \cdot r = I \cdot \frac{a}{r}
\]
Substituting \( I \):
\[
T \cdot r = \left(\frac{1}{2} m_p r^2\right) \cdot \frac{a}{r}
\]
This simplifies to:
\[
T = \frac{1}{2} m_p a
\]

### Step 4: Substitute into the force equation for the hanging mass
Now substituting \( T \) into the first equation:
\[
mg - \frac{1}{2} m_p a = ma
\]
Rearranging gives:
\[
mg = ma + \frac{1}{2} m_p a
\]
Factoring out \( a \):
\[
mg = a \left( m + \frac{1}{2} m_p \right)
\]
Solving for \( a \):
\[
a = \frac{mg}{m + \frac{1}{2} m_p}
\]

### Step 5: Calculate \( a \)
Substituting the known values:
\[
a = \frac{0.270 \cdot 9.81}{0.270 + \frac{1}{2} \cdot 1.50}
\]
Calculating:
\[
a = \frac{2.6457}{0.270 + 0.75} = \frac{2.6457}{1.020} \approx 2.594 \, \text{m/s}^2
\]

### Step 6: Calculate the tension \( T \)
Now substituting back to find \( T \):
\[
T = \frac{1}{2} m_p a = \frac{1}{2} \cdot 1.50 \cdot 2.594 \approx 1.938 \, \text{N}
\]

### Final Answer
The tension in the string is approximately:
\[
\boxed{1.94 \, \text{N}}
\]

Do you want further details on any step or have any questions? Here are some related questions you might consider:

1. How would the tension change if the mass of the hanging object increased?
2. What would happen to the acceleration if the radius of the pulley was smaller?
3. How does the moment of inertia affect the rotational motion of the pulley?
4. If the pulley were frictionless, how would that affect the tension and acceleration?
5. What assumptions were made in this analysis regarding the string and pulley? 

Tip: Always ensure to check unit consistency when performing calculations.

A solid cylindrical pulley with a mass of m=1.50 kg and a radius of r=0.270 m is free to rotate about its axis. An object of mass 0.270 kg is attached to the pulley with a light string. Calculate the tension in the string.

Explore how to calculate the tension in a string attached to a rotating pulley with a mass of 1.50 kg and radius 0.270 m, supporting a 0.270 kg mass. This detailed solution uses concepts from dynamics and rotational motion.

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