How Do You Calculate Time from Angular Velocity and Arc Length?

[Lfrizz]

The radius of the circle traced out by the second hand on a clock is 6.00 cm. In a time t the tip of the second hand moves through an arc length of 19.0 cm. Determine the value of t in seconds.


Ok, so I know how to find θ: s/r. And I know that Angular Velocity is θ/Δt. I am confused on how I get the angular velocity so I can find the change in time. Also, do I need to convert anytihng into radians, or is it already in radians since it's the RADIUS of the circle... I feel really dumb asking this since I know it is so simple, but I am confused.

Thanks.
Leah

 

[Lfrizz]

I think I might be on to something...

the t for the entire radius should be 60 seconds... so would I divide .06m/60s to get the avg. angular velocity? Then I could use that with the theta I find for the arc length of .19m to find the time for that distance?

 

[cupid.callin]

I think I might be on to something...

the t for the entire radius should be 60 seconds... so would I divide .06m/60s to get the avg. angular velocity? Then I could use that with the theta I find for the arc length of .19m to find the time for that distance?

Your initial thinking was correct but you proceeded wrongly ...

T = 60s is time when it has covered a full revolution ... what do you think will be θ for that??

then use w = θ/T

 

[Lfrizz]

I think it would = 2[tex]\pi[/tex]r.

So... I will use the formula w = θ/T
Where [tex]\theta[/tex]= 2[tex]\pi[/tex]r and r=0.19m
divided by T=60s
to find omega [tex]\omega[/tex] (I'm still trying to get familiar with the little reference list on the side, I'm going blind)

Then I use this Omega with the theta I found above and find the new time?

 

[Lfrizz]

I messed up. r=.06 not 0.19.

 

[cupid.callin]

θ is angular distance ... it must be in radians

if your distance traveled about some point is 2πr ... then angular distance is 2π

remember s=θr ?

 

[Lfrizz]

I'm still confused how I find the T of just that one small portion of the rotation. Sorry if I seem dense, I read to much into these problems. And thank you for your help!

T for the second hand to make an entire rotation is 60 sec
the radius of the entire rotation is .06m
the arc length is .19m
... [tex]\theta[/tex] = .19/.06= 3.16 radians

I'm not sure where the 60 seconds comes in now. I have thoroughly confused myself with this basic problem.

 

[Lfrizz]

I think I might get what your saying...

2[tex]\pi[/tex]/60 would give me [tex]\omega[/tex]

then I can plug in 3.16 from my previous post for [tex]\theta[/tex] and solve for T ?

 

[Lfrizz]

It says I got it right... Wow Thanks!
[tex]\omega[/tex]=2[tex]\pi[/tex]/60sec = .105
w = θ/T
.105=3.16/T
T= 30.09


THANK YOU!

 

[cupid.callin]

2[tex]\pi[/tex]/60 would give me [tex]\omega[/tex]

Yes

and for a complete rotation you know arc length(s) is 2πr (r is radius)

so θ = s/r = 2πr/r = 2π (which is 3.14)

then use the w to find speed(v) of tip hand

v = wr

then find t using v

there is a simpler way of doing this problem but as you started with w so i told you how to solve using w

 

[Lfrizz]

What is the easier way? lol

 

[cupid.callin]

ok ... so let radius is R
and distance traveled is d

consider tip of second hand ...
lets find its speed

v = d/t

so v = 2πR / (60 sec)

v found

now you have distance .19

so .19 = v * (t_required)

 

[Lfrizz]

that makes more sense.