Time on Planet X: Solving for the Alignment of Clock Hands

[phyhelp]

Homework Statement

A wall clock on Planet X has two hands that are aligned at midnight and turn in the same direction at uniform rates, one at 4.27×10−2 rad/ s and the other at 1.62×10−2 rad/s . At how many seconds after midnight are these hands first aligned and next aligned?

Homework Equations

ω= θ/ t

The Attempt at a Solution

I don't know where to start. Should i try to solve for when their angles are equal?

 

[Sleepy_time]

Hi phyhelp. Think about how much of an angle the faster hand has to go through to meet up with the slower hand for the second time, the first time, their angles are 0 at t=0. Substitute your equation [itex]\omega=\frac{\theta}{t}[/itex] for each angle with the appropriate angular frequency and solve for t.

 

[azizlwl]

Should i try to solve for when their angles are equal?
Yes.
Find the time the faster hand to turn 1 rev or 2∏ radians.
Then start chasing the slower hand which in turn has moved to certain angle from midnight.

 

[Sleepy_time]

No, not equal, that would just give t=0. Solve for when the angle is as you said 2[itex]\pi[/itex] of the other one. For the second time they meet after t=0, this would just be 2revolutions or 4[itex]\pi[/itex]. You can see that in general they'll meet when [itex]\theta_1=\theta_2+2n\pi[/itex], where n is the n'th time they meet after t=0.

 

[phyhelp]

I'm sorry but I'm still lost. Using the equation ω=θ/t, i solved for t when θ= 2∏ rad and ω= 4.27*10^-2 and 1.62*10^-2 and I got 147.147 s and 387.85 s, respectively. Am I on the right track?

 

[Sleepy_time]

What you're working out there is the time it takes for the respective hand to cover an angle of 2[itex]\pi[/itex]. Instead use [itex]\theta_1=\theta_2+2\pi[/itex], [itex]\omega_1=\frac{\theta_1}{t}[/itex] & [itex]\omega_2=\frac{\theta_2}{t}[/itex], where 1 is the faster hand and 2 the other.

 

[azizlwl]

ω1 the faster one.
Time for ω1 to return to midnight position is also the time for ω2 to move to θ1
Now start the race. ω1 at midnight and ω2 at angle θ1 ahead.
They will meet at same angle θ2 at the same time, t

ω1t=θ2 (1)
ω2t+θ1=θ2 (2)
or
ω2t+T_ω1ω2=θ2

Then at θ2 is your new second virtual midnight.

 

[Villyer]

Another option for looking at this problem is by saying that the difference of the two hands angular positions is a multiple of 2π.
So you can say that t(ω₁ - ω₂) = 2π * n
If n = 0 then t = 0, which means that the hands line up at t = 0 (which we are told).
So if you set n = 1, it will give the first time that the hands line up again.

 

[phyhelp]

Thank you everyone for all your help!