Math Problem Statement

The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. The bob has a mass of 0.0420 kg, the string has length L = 1.04 m and negligible mass, and the bob follows a circular path of circumference 1.25 m. What are (a) the tension in the string and (b) the period of the motion?

Solution

The problem you've uploaded involves a conical pendulum, where a bob moves in a horizontal circle at constant speed. Here's the breakdown:

Given Data:

  • Mass of the bob: m=0.0420kgm = 0.0420 \, \text{kg}
  • Length of the string: L=1.04mL = 1.04 \, \text{m}
  • Circumference of the circular path: C=1.25mC = 1.25 \, \text{m}

You are tasked with finding:

  1. (a) The tension in the string.
  2. (b) The period of the motion.

Step 1: Calculate the Radius of the Circular Path

The radius rr of the circular path can be calculated from the given circumference: C=2πrr=C2πC = 2\pi r \quad \Rightarrow \quad r = \frac{C}{2\pi} Substitute C=1.25mC = 1.25 \, \text{m}: r=1.252π=0.199mr = \frac{1.25}{2\pi} = 0.199 \, \text{m}

Step 2: Relate Geometry to Tension

The angle θ\theta between the string and the vertical can be found using the geometry of the conical pendulum: cosθ=hL\cos \theta = \frac{h}{L} where hh is the vertical height of the string, and r=Lsinθr = L \sin \theta. This gives: sinθ=rL=0.1991.04=0.191\sin \theta = \frac{r}{L} = \frac{0.199}{1.04} = 0.191

Now find θ\theta: θ=sin1(0.191)11.01\theta = \sin^{-1}(0.191) \approx 11.01^\circ

Step 3: Tension in the String

To calculate the tension TT, we consider the forces acting on the bob. The tension provides both a centripetal force Tsinθ=mv2rT \sin \theta = \frac{mv^2}{r} and supports the weight of the bob Tcosθ=mgT \cos \theta = mg.

We can find TT by solving the system of equations. The vertical component balances the weight: Tcosθ=mgT=mgcosθT \cos \theta = mg \quad \Rightarrow \quad T = \frac{mg}{\cos \theta} Substitute m=0.0420kg,g=9.8m/s2,cosθ=cos11.01=0.981m = 0.0420 \, \text{kg}, \, g = 9.8 \, \text{m/s}^2, \, \cos \theta = \cos 11.01^\circ = 0.981: T=0.0420×9.80.9810.419NT = \frac{0.0420 \times 9.8}{0.981} \approx 0.419 \, \text{N}

Step 4: Period of the Motion

The period TperiodT_{\text{period}} can be found by using the velocity and the radius. First, calculate the velocity vv. Using Tsinθ=mv2rT \sin \theta = \frac{mv^2}{r}, we solve for vv: v=Tsinθrmv = \sqrt{\frac{T \sin \theta \cdot r}{m}} Substitute T=0.419N,sinθ=0.191,r=0.199m,m=0.0420kgT = 0.419 \, \text{N}, \sin \theta = 0.191, r = 0.199 \, \text{m}, m = 0.0420 \, \text{kg}: v=0.419×0.191×0.1990.04200.58m/sv = \sqrt{\frac{0.419 \times 0.191 \times 0.199}{0.0420}} \approx 0.58 \, \text{m/s}

The period TperiodT_{\text{period}} is: Tperiod=2πrvT_{\text{period}} = \frac{2\pi r}{v} Substitute r=0.199m,v=0.58m/sr = 0.199 \, \text{m}, v = 0.58 \, \text{m/s}: Tperiod=2π×0.1990.582.16secondsT_{\text{period}} = \frac{2\pi \times 0.199}{0.58} \approx 2.16 \, \text{seconds}

Final Answers:

  1. Tension in the string: T0.419NT \approx 0.419 \, \text{N}
  2. Period of motion: Tperiod2.16secondsT_{\text{period}} \approx 2.16 \, \text{seconds}

Would you like further clarifications or more details on any step?

Here are 5 related questions for further exploration:

  1. How would the tension change if the mass of the bob were doubled?
  2. How does the period of motion depend on the radius of the circular path?
  3. What would happen to the period if the length of the string increased?
  4. How would the tension change if the bob moved faster in the circular path?
  5. What effect does gravitational acceleration have on the period?

Tip: Always break problems involving forces into their horizontal and vertical components for clarity in solving.

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Math Problem Analysis

Mathematical Concepts

Circular motion
Trigonometry
Centripetal force
Newton's Second Law

Formulas

T = mg / cos(θ)
v = sqrt(T sin(θ) * r / m)
T_period = 2πr / v
r = C / (2π)

Theorems

Newton's Second Law of Motion
Centripetal force equation

Suitable Grade Level

Grades 11-12