Question

At 3:00 the hour hand and the minute hand of a clock point indirections that are 90.0 degrees apart. What is the firsttime after 3:00 that the angle between the two hands has decreasedby half to 45.0 degrees?

Answer 1

For the angle between the minute hand and the hour hand todecrease by 45^o, the minute hand must rotate through45^o relative to the hour hand. Both are moving in theclockwise direction - the min hand faster.

The angular speeds are, ω₁ = 2π rad/ 60min, for the min hand and ω₂ = 2π rad/12 hour,for the hour hand.

(each takes 60 min and 12 hour respectively to make onecomplete rotation or 2π radians.).

Rel speed of the min hand with respect to the hourhand,

            ω = ω₁ - ω₂ =(2π/60)(1 - 1/12) = 11π/360 rad/min.

At this speed it must cover, θ=  45^o = π/4 rad.

Using     θ = ωt

Time taken,

            t = θ/ω = (π/4)/11π/360 = 90/11

               = 8.1818 min = 8.18min.

This is at time3:08:11 (8 min and 11 s past 3:00).

Answer 2

For theangle between the minute hand and the hour hand to decrease by45^o, the minute hand must rotate through 45^orelative to the hour hand. Both are moving in the clockwisedirection - the min hand faster.

Theangular speeds are, ω₁ = 2π rad/ 60 min, forthe min hand and ω₂ = 2π rad/12 hour, for thehour hand.

(eachtakes 60 min and 12 hour respectively to make one complete rotationor 2π radians.).

Relspeed of the min hand with respect to the hourhand,

            ω = ω₁ - ω₂ =(2π/60)(1 - 1/12) = 11π/360 rad/min.

At thisspeed it must cover, θ =  45^o = π/4rad.

Using    θ = ωt

Timetaken,

            t = θ/ω = (π/4)/11π/360 =90/11

               = 8.1818 min = 8.18min.

This isat time 3:08:11 (8min and 11 s past 3:00).

I do not understand how you got the relative speed of the minutehand with respect to the hour hand. How do you know to useω= ω1-ω2?