Angular and Linear Speed....Part 1

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SUMMARY

The discussion focuses on calculating angular and linear speeds based on a wheel's rotation rate and radius. Given a rotation rate of 500 rpm and a radius of 45 cm, the correct angular speed is determined to be 50π/3 radians/sec. The linear speed at the circumference is calculated as 750π cm/sec, while the linear speed halfway to the center is 375π cm/sec. The discrepancies in calculations arise from the need to convert units appropriately, particularly from minutes to seconds.

PREREQUISITES
  • Understanding of angular speed and linear speed concepts
  • Familiarity with the formulas for angular speed (ω = θ/t) and linear speed (v = ωr)
  • Knowledge of unit conversions, specifically between revolutions per minute and radians per second
  • Basic proficiency in trigonometric functions and their applications in physics
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  • Study unit conversion techniques, particularly between rpm and radians/sec
  • Learn about the relationship between angular and linear velocity in rotational motion
  • Explore the implications of radius changes on linear speed calculations
  • Practice solving problems involving angular motion using different radius values
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Students studying physics, educators teaching rotational dynamics, and anyone interested in understanding the principles of angular and linear motion calculations.

mathdad
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You are given the rate of rotation of a wheel as well as its radius. For A-C, determine the following:

A. The angular speed, in units of radians/sec.

B. The linear speed, in units of cm/sec, of a point on the circumference of the wheel.

C. The linear speed, in cm/sec, of a point halfway between the center of the wheel and the circumference.

I will now post my effort.

Given: 500 rpm; r = 45 cm

Part A

The angular speed formula is w = θ/t.

I know that 1 revolution = 2pi radians.

I need θ.

θ = 500 (2pi radians)

θ = 1,000 pi radians.

w = 1,000 pi radians/sec

Book's answer is [50 pi/3] radians/sec. Something tells me that I needed to convert seconds to minutes. Yes?

Part B

I used the arc length formula, s = θr, as step 1.

s = (1,000 pi radians)(45 cm)

s = 45,000 pi cm = d

The letter d represents the distance in time t in the linear speed formula v = d/t.

v = 45,000 pi cm/sec is my answer.

Book's answer for Part B is 750 pi cm/sec.
Again, I am thinking that the units of conversation needed to be changed. Yes?

Part C

s = θr

s = (1,000 pi radians)(22.5 cm)

The decimal 22.5 came from dividing the radius in half in terms of the instructions for Part C above.

s = 22,500 pi cm = d

v = d/t

v = 22,500 pi cm/sec

Book's answer is 375 pi cm/sec.
 
Last edited:
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A. $$\omega=\frac{\theta}{t}$$

$$\omega=\frac{500\text{ rev}}{1\text{ min}}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}}\cdot\frac{1\text{ min}}{60\text{ s}}=\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}$$

B. $$v=\omega r$$

$$v=\left(\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}\right)\left(45\text{ cm}\right)=750\pi\frac{\text{cm}}{\text{s}}$$

C. $$v=\omega r$$

$$v=\left(\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}\right)\left(\frac{45}{2}\text{ cm}\right)=375\pi\frac{\text{cm}}{\text{s}}$$
 
MarkFL said:
A. $$\omega=\frac{\theta}{t}$$

$$\omega=\frac{500\text{ rev}}{1\text{ min}}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}}\cdot\frac{1\text{ min}}{60\text{ s}}=\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}$$

B. $$v=\omega r$$

$$v=\left(\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}\right)\left(45\text{ cm}\right)=750\pi\frac{\text{cm}}{\text{s}}$$

C. $$v=\omega r$$

$$v=\left(\frac{50\pi}{3}\,\frac{\text{rad}}{\text{s}}\right)\left(\frac{45}{2}\text{ cm}\right)=375\pi\frac{\text{cm}}{\text{s}}$$

I follow everything you did here except for Part A.
Where did 50 pi/3 come from? Don't we multiply 500 rpm by 2 pi radians? Is 500 rpm • 2 pi radians not equal to 1,000 pi radians?
 
RTCNTC said:
I follow everything you did here except for Part A.
Where did 50 pi/3 come from? Don't we multiply 500 rpm by 2 pi radians? Is 500 rpm • 2 pi radians not equal to 1,000 pi radians?

Yes, but we also must divide by 60, and then reduce the fraction. :)
 
Oh boy! I better go back to 5th grade math.
 

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