Question

A penny is placed a distance r from the center of a record spinning at ω = 90π rad/min. The coefficient of static friction between the penny and the record is μs = 0.14 on the horizontal plane.

What is the distance, r in meters?

Answer 1

Well, I am here.

Answer 2

The equation to find r is \[(\mu*g)/\omega^2\] but I'm not getting the right answer for some reason

Answer 3

Where are you stuck?

Answer 4

getting the right answer

Answer 5

\[\frac{ \mu*g }{ \omega^2 } =??\]

Answer 6

You need to be careful with the units

Answer 7

So this is how I did it.. \[\frac{ 0.14 *9.81}{ 90\pi^2}\]

Answer 8

that denominator is wrong

Answer 9

Hint:
1. convert min to sec.
2. square the w>>=\(w^2=90^2*pi^2\)

Answer 10

but that without conversion to rad/sec

Answer 11

9.81 is in m/s-2 and speed is in in minutes

Answer 12

convert to rads/sec

Answer 13

@simplixity

Answer 14

soo would I do this? \[\frac{ 90\pi }{ 1 rev }*\frac{ 1 \min }{ 60 \sec }\] to convert?

Answer 15

I got 4.712

Answer 16

Great!

Sub. with it it the formula you typed!

Answer 17

is this how it should look? \[\frac{ 0.14*4.712 }{ 90^2\pi^2 }\]

Answer 18

I am here!

Answer 19

No, I don't think so.

It would be\[\Large \frac{ \mu*g }{ \omega^2 } =\frac{ (0.14)(9.81) }{ (4.7124)^2 }=0.06185\]

I ask you know, what does this value represent?

Answer 20

Is there any graph for that problem?

Answer 21

thank you for your help

Answer 22

Thank you for the medal!

Did you get what you were looking for?

Answer 23

no problem and yes, I understand it now

Answer 24

Any more questions?

Answer 25

I am happy to hear that.

Answer 26

nope that's it. thank you, I appreciate it

Answer 27

I will not be late for any help.