Math Problem Statement

A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. What is angular speed (in rpm ) or a 500 g ball swinging in a 20- cm -radius circle at the end of a 1.0- m -long string?

Solution

We can calculate the angular speed of the conical pendulum using the following steps:

Step 1: Variables and Definitions

Given:

  • Mass of the ball, m=500g=0.5kgm = 500 \, \text{g} = 0.5 \, \text{kg}
  • Radius of the circular motion, r=20cm=0.2mr = 20 \, \text{cm} = 0.2 \, \text{m}
  • Length of the string, L=1.0mL = 1.0 \, \text{m}

Step 2: Geometry of the Conical Pendulum

For a conical pendulum, the string makes an angle θ\theta with the vertical. This angle can be determined from the relation: sinθ=rL\sin \theta = \frac{r}{L} Substituting the values of rr and LL: sinθ=0.21.0=0.2\sin \theta = \frac{0.2}{1.0} = 0.2 Thus, θ=arcsin(0.2)\theta = \arcsin(0.2).

Step 3: Dynamics of the Conical Pendulum

The forces acting on the pendulum are:

  • Tension in the string, TT
  • Gravitational force, mgmg

In equilibrium, the vertical component of tension balances the gravitational force: Tcosθ=mgT \cos \theta = mg The horizontal component provides the centripetal force: Tsinθ=mv2rT \sin \theta = m \frac{v^2}{r} Where vv is the linear speed of the ball. Using these two equations, we can eliminate TT and solve for vv.

Dividing the second equation by the first: tanθ=v2rg\tan \theta = \frac{v^2}{rg} Thus, the linear speed vv is: v=rgtanθv = \sqrt{r g \tan \theta} Substitute r=0.2mr = 0.2 \, \text{m}, g=9.8m/s2g = 9.8 \, \text{m/s}^2, and tanθ=0.2027\tan \theta = 0.2027 (from θ=arcsin(0.2)\theta = \arcsin(0.2)): v=0.2×9.8×0.20270.632m/sv = \sqrt{0.2 \times 9.8 \times 0.2027} \approx 0.632 \, \text{m/s}

Step 4: Angular Speed Calculation

The angular speed ω\omega is related to the linear speed vv by: ω=vr\omega = \frac{v}{r} Substituting v=0.632m/sv = 0.632 \, \text{m/s} and r=0.2mr = 0.2 \, \text{m}: ω=0.6320.23.16rad/s\omega = \frac{0.632}{0.2} \approx 3.16 \, \text{rad/s}

Step 5: Conversion to RPM

To convert ω\omega from radians per second to revolutions per minute (rpm), use the following conversion: 1rad/s=602πrpm1 \, \text{rad/s} = \frac{60}{2\pi} \, \text{rpm} Thus, ω3.16×602π30.2rpm\omega \approx 3.16 \times \frac{60}{2\pi} \approx 30.2 \, \text{rpm}

Final Answer:

The angular speed of the ball is approximately 30.2 rpm.

Would you like further clarification or details on any part of the process? Here are five related questions for exploration:

  1. How does the length of the string affect the angular speed?
  2. What would happen if the mass of the ball was increased while keeping the radius the same?
  3. How would the angular speed change if the pendulum was swinging at a larger radius?
  4. Can you derive the time period of a conical pendulum?
  5. How is the angle θ\theta affected by changes in the radius of the motion?

Tip: Remember that angular speed depends on both the radius and the gravitational force, so any change in these factors will alter the pendulum's dynamics.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Trigonometry
Centripetal Force
Angular Speed

Formulas

sin(θ) = r / L
Tcos(θ) = mg
Tsin(θ) = mv^2 / r
v = sqrt(r * g * tan(θ))
ω = v / r
1 rad/s = 60 / 2π rpm

Theorems

Newton's Second Law
Trigonometric Relationships

Suitable Grade Level

Grades 11-12