Math Problem Statement

A circular disk whose radius is 3 m is placed on a horizontal floor and is moving with an angular velocity governed by the equation ω = 2t^2 + 3 rad/sec. Determine the values of radial acceleration, tangential acceleration, and the net acceleration of that circular disk at time 3.00 sec.

Solution

To solve the given problem, let’s break it into steps.

Problem Details:

  • Radius of the circular disk: r=3mr = 3 \, \text{m}
  • Angular velocity: ω=2t2+3rad/s\omega = 2t^2 + 3 \, \text{rad/s}
  • Time of interest: t=3st = 3 \, \text{s}
  • Quantities to find:
    1. Radial acceleration (ara_r)
    2. Tangential acceleration (ata_t)
    3. Net acceleration (aneta_{\text{net}})

Step 1: Angular velocity (ω\omega) at t=3st = 3 \, \text{s}

The angular velocity is given by: ω=2t2+3\omega = 2t^2 + 3 Substitute t=3t = 3: ω=2(3)2+3=2(9)+3=18+3=21rad/s\omega = 2(3)^2 + 3 = 2(9) + 3 = 18 + 3 = 21 \, \text{rad/s}


Step 2: Angular acceleration (α\alpha)

The angular acceleration is the time derivative of angular velocity: α=dωdt\alpha = \frac{d\omega}{dt} From ω=2t2+3\omega = 2t^2 + 3, differentiate: α=ddt(2t2+3)=4t\alpha = \frac{d}{dt}(2t^2 + 3) = 4t Substitute t=3t = 3: α=4(3)=12rad/s2\alpha = 4(3) = 12 \, \text{rad/s}^2


Step 3: Tangential acceleration (ata_t)

The tangential acceleration is related to angular acceleration: at=αra_t = \alpha r Substitute α=12rad/s2\alpha = 12 \, \text{rad/s}^2 and r=3mr = 3 \, \text{m}: at=12(3)=36m/s2a_t = 12(3) = 36 \, \text{m/s}^2


Step 4: Radial acceleration (ara_r)

The radial acceleration is given by: ar=ω2ra_r = \omega^2 r Substitute ω=21rad/s\omega = 21 \, \text{rad/s} and r=3mr = 3 \, \text{m}: ar=(21)2(3)=441(3)=1323m/s2a_r = (21)^2(3) = 441(3) = 1323 \, \text{m/s}^2


Step 5: Net acceleration (aneta_{\text{net}})

The net acceleration is the vector sum of radial and tangential accelerations: anet=ar2+at2a_{\text{net}} = \sqrt{a_r^2 + a_t^2} Substitute ar=1323m/s2a_r = 1323 \, \text{m/s}^2 and at=36m/s2a_t = 36 \, \text{m/s}^2: anet=(1323)2+(36)2a_{\text{net}} = \sqrt{(1323)^2 + (36)^2} First, calculate the squares: (1323)2=1750329,(36)2=1296(1323)^2 = 1750329, \quad (36)^2 = 1296 Add the results: anet=1750329+1296=1751625a_{\text{net}} = \sqrt{1750329 + 1296} = \sqrt{1751625} Take the square root: anet1323.49m/s2a_{\text{net}} \approx 1323.49 \, \text{m/s}^2


Final Answers:

  1. Radial acceleration: ar=1323m/s2a_r = 1323 \, \text{m/s}^2
  2. Tangential acceleration: at=36m/s2a_t = 36 \, \text{m/s}^2
  3. Net acceleration: anet1323.49m/s2a_{\text{net}} \approx 1323.49 \, \text{m/s}^2

Would you like further clarification or more detailed steps? Here are related questions to expand your understanding:

  1. What is the significance of radial acceleration in circular motion?
  2. How is angular acceleration derived from the angular velocity equation?
  3. What is the relationship between angular and tangential acceleration?
  4. How does the radius of the disk affect the radial and tangential accelerations?
  5. Can we calculate the velocity of a point on the disk's edge at t=3st = 3 \, \text{s}?

Tip: Always differentiate carefully when dealing with time-dependent quantities like angular velocity or position to avoid errors in acceleration calculations.

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Math Problem Analysis

Mathematical Concepts

Circular motion
Angular velocity
Radial acceleration
Tangential acceleration
Net acceleration

Formulas

Angular velocity: ω = 2t^2 + 3
Angular acceleration: α = dω/dt
Tangential acceleration: at = αr
Radial acceleration: ar = ω^2r
Net acceleration: anet = √(ar^2 + at^2)

Theorems

Laws of circular motion

Suitable Grade Level

Grades 11-12