Math Problem Statement

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.)

Solution

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, mp=1.50kgm_p = 1.50 \, \text{kg}
  • Radius of the pulley, r=0.270mr = 0.270 \, \text{m}
  • Mass of the hanging object, m=0.270kgm = 0.270 \, \text{kg}
  • Acceleration due to gravity, g=9.81m/s2g = 9.81 \, \text{m/s}^2

Step 1: Analyze the forces on the hanging mass

For the hanging mass mm: Fnet=mgT=maF_{\text{net}} = mg - T = ma where TT is the tension in the string and aa is the linear acceleration of the mass.

Step 2: Analyze the pulley

The torque τ\tau on the pulley due to the tension in the string is given by: τ=Tr\tau = T \cdot r The torque is also related to the angular acceleration α\alpha of the pulley: τ=Iα\tau = I \alpha where II is the moment of inertia of the pulley. For a solid cylinder: I=12mpr2I = \frac{1}{2} m_p r^2 The relationship between linear acceleration aa and angular acceleration α\alpha is: a=rαα=ara = r \alpha \quad \Rightarrow \quad \alpha = \frac{a}{r}

Step 3: Substitute into the torque equation

Substituting for α\alpha: Tr=IarT \cdot r = I \cdot \frac{a}{r} Substituting II: Tr=(12mpr2)arT \cdot r = \left(\frac{1}{2} m_p r^2\right) \cdot \frac{a}{r} This simplifies to: T=12mpaT = \frac{1}{2} m_p a

Step 4: Substitute into the force equation for the hanging mass

Now substituting TT into the first equation: mg12mpa=mamg - \frac{1}{2} m_p a = ma Rearranging gives: mg=ma+12mpamg = ma + \frac{1}{2} m_p a Factoring out aa: mg=a(m+12mp)mg = a \left( m + \frac{1}{2} m_p \right) Solving for aa: a=mgm+12mpa = \frac{mg}{m + \frac{1}{2} m_p}

Step 5: Calculate aa

Substituting the known values: a=0.2709.810.270+121.50a = \frac{0.270 \cdot 9.81}{0.270 + \frac{1}{2} \cdot 1.50} Calculating: a=2.64570.270+0.75=2.64571.0202.594m/s2a = \frac{2.6457}{0.270 + 0.75} = \frac{2.6457}{1.020} \approx 2.594 \, \text{m/s}^2

Step 6: Calculate the tension TT

Now substituting back to find TT: T=12mpa=121.502.5941.938NT = \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: 1.94N\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.

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Math Problem Analysis

Mathematical Concepts

Dynamics
Torque
Rotational Motion

Formulas

F_net = mg - T = ma
τ = T * r
τ = I * α
I = (1/2) * m_p * r^2
a = r * α

Theorems

Newton's Second Law
Torque and Angular Acceleration

Suitable Grade Level

Grades 10-12