Math Problem Statement

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.280 rev/s . The magnitude of the angular acceleration is 0.916 rev/s2 . Both the the angular velocity and angular accleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.700 m . What is the tangential speed vt of a point on the tip of the blade at time t = 0.197 s ?

Solution

To find the tangential speed vtv_t of a point on the tip of the blade at time t=0.197st = 0.197 \, \text{s}, we need to use the relationship between angular speed and tangential speed.

The tangential speed vtv_t at a point on the blade is given by: vt=ωrv_t = \omega \cdot r where:

  • ω\omega is the angular velocity in radians per second,
  • rr is the radius of the fan blade.

Let's break the solution into steps:

Step 1: Convert angular quantities to standard units

We are given:

  • Initial angular velocity: ω0=0.280rev/s\omega_0 = 0.280 \, \text{rev/s},
  • Angular acceleration: α=0.916rev/s2\alpha = 0.916 \, \text{rev/s}^2.

Since we need angular velocity in radians per second, we can use the conversion: 1rev=2πradians1 \, \text{rev} = 2\pi \, \text{radians} Thus: ω0=0.280×2πrad/s=1.7593rad/s\omega_0 = 0.280 \times 2\pi \, \text{rad/s} = 1.7593 \, \text{rad/s} α=0.916×2πrad/s2=5.7545rad/s2\alpha = 0.916 \times 2\pi \, \text{rad/s}^2 = 5.7545 \, \text{rad/s}^2

Step 2: Find angular velocity at t=0.197st = 0.197 \, \text{s}

The angular velocity ω\omega at time tt can be found using the equation: ω=ω0+αt\omega = \omega_0 + \alpha t Substituting the known values: ω=1.7593+(5.7545×0.197)\omega = 1.7593 + (5.7545 \times 0.197) ω=1.7593+1.1342=2.8935rad/s\omega = 1.7593 + 1.1342 = 2.8935 \, \text{rad/s}

Step 3: Find the radius of the blade

The radius rr is half of the diameter. Given the diameter of the fan blade is 0.700m0.700 \, \text{m}, we have: r=0.7002=0.350mr = \frac{0.700}{2} = 0.350 \, \text{m}

Step 4: Calculate the tangential speed

Now, using the formula vt=ωrv_t = \omega \cdot r: vt=2.8935×0.350=1.0127m/sv_t = 2.8935 \times 0.350 = 1.0127 \, \text{m/s}

Thus, the tangential speed at the tip of the blade at time t=0.197st = 0.197 \, \text{s} is approximately: vt1.01m/sv_t \approx 1.01 \, \text{m/s}

Would you like more details on any part of this solution, or have any questions?


Here are 5 related questions:

  1. What would be the tangential speed at t=1st = 1 \, \text{s}?
  2. How would the tangential speed change if the angular acceleration were doubled?
  3. What is the centripetal acceleration of the point on the blade tip at t=0.197st = 0.197 \, \text{s}?
  4. How long would it take for the fan to reach a tangential speed of 2m/s2 \, \text{m/s}?
  5. How many revolutions does the fan complete in 0.197s0.197 \, \text{s}?

Tip: Always check units to ensure consistency when solving physics problems involving motion.

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Math Problem Analysis

Mathematical Concepts

Angular velocity
Tangential speed
Circular motion

Formulas

v_t = ω × r
ω = ω_0 + α × t
1 rev = 2π radians

Theorems

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Suitable Grade Level

Grades 10-12