Angular and Linear Speed....Part 1

[mathdad]

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.

 

[MarkFL]

A. \(\displaystyle \omega=\frac{\theta}{t}\)

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

B. \(\displaystyle v=\omega r\)

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

C. \(\displaystyle v=\omega r\)

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

 

[mathdad]

A. \(\displaystyle \omega=\frac{\theta}{t}\)

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

B. \(\displaystyle v=\omega r\)

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

C. \(\displaystyle v=\omega r\)

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

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?

 

[MarkFL]

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. :)

 

[mathdad]

Oh boy! I better go back to 5th grade math.