Math Problem Statement

A 0.60-kg rock is swung in a circular path and in a vertical plane on a 0.25-m length string. At the top of the path, the angular speed is 12.0 rad/s. What is the tension in the string at that point?

Solution

To find the tension in the string at the top of the circular path, we can use the following principles of circular motion:

  1. Centripetal Force: This is the force required to keep the rock moving in a circular path. At the top of the path, the centripetal force is provided by the tension in the string and the weight of the rock.

  2. Formula for Centripetal Force: Fcentripetal=mv2rF_{\text{centripetal}} = m \cdot \frac{v^2}{r} where:

    • mm is the mass of the rock,
    • vv is the linear speed of the rock, and
    • rr is the radius of the circular path.
  3. Relating Angular Speed and Linear Speed: The linear speed vv is related to the angular speed ω\omega by the formula: v=ωrv = \omega \cdot r

  4. Tension Calculation at the Top of the Path: At the top, the tension TT in the string and the gravitational force mgmg both act downward, providing the centripetal force. Thus, we have: T+mg=mv2rT + mg = m \cdot \frac{v^2}{r}

Now, let’s calculate each part:

Step 1: Given Values

  • Mass of rock, m=0.60kgm = 0.60 \, \text{kg}
  • Radius of circular path, r=0.25mr = 0.25 \, \text{m}
  • Angular speed, ω=12.0rad/s\omega = 12.0 \, \text{rad/s}
  • Gravitational acceleration, g=9.81m/s2g = 9.81 \, \text{m/s}^2

Step 2: Find the Linear Speed vv

Using the formula v=ωrv = \omega \cdot r: v=12.0rad/s×0.25m=3.0m/sv = 12.0 \, \text{rad/s} \times 0.25 \, \text{m} = 3.0 \, \text{m/s}

Step 3: Calculate Centripetal Force

Now, substitute vv into the centripetal force formula: Fcentripetal=mv2r=0.60kg(3.0m/s)20.25mF_{\text{centripetal}} = m \cdot \frac{v^2}{r} = 0.60 \, \text{kg} \cdot \frac{(3.0 \, \text{m/s})^2}{0.25 \, \text{m}} Fcentripetal=0.60kg9.0m2/s20.25m=0.60×36=21.6NF_{\text{centripetal}} = 0.60 \, \text{kg} \cdot \frac{9.0 \, \text{m}^2/\text{s}^2}{0.25 \, \text{m}} = 0.60 \times 36 = 21.6 \, \text{N}

Step 4: Solve for Tension TT

Using T+mg=FcentripetalT + mg = F_{\text{centripetal}}, rearrange to solve for TT: T=FcentripetalmgT = F_{\text{centripetal}} - mg T=21.6N(0.60kg×9.81m/s2)T = 21.6 \, \text{N} - (0.60 \, \text{kg} \times 9.81 \, \text{m/s}^2) T=21.6N5.886N=15.714NT = 21.6 \, \text{N} - 5.886 \, \text{N} = 15.714 \, \text{N}

Final Answer

The tension in the string at the top of the path is approximately: T15.7NT \approx 15.7 \, \text{N}

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

  1. How would the tension change if the angular speed were increased?
  2. What happens to the tension if the string length is doubled?
  3. How does the tension vary as the rock moves through different points in the vertical circle?
  4. What role does gravitational force play in this scenario?
  5. How would the calculations change if the rock were swung in a horizontal plane instead?

Tip: When solving circular motion problems, always account for the direction of forces, especially gravity, which may either add to or reduce the required tension depending on the point in the motion.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Circular Motion
Centripetal Force
Gravitational Force
Angular Speed

Formulas

Centripetal Force: F_c = m * v^2 / r
Linear Speed: v = ω * r
Tension at the top: T + mg = F_c

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12