Whiplash!

Consider a rope attached to one vertex of a regular hexagon, and originally resting at right angles with the symmetry line through that vertex. The rope moves towards the edge with a certain angular velocity. Once it comes in contact with the first edge, the rope bends around the first edge, and moves with the same linear velocity towards the second edge. Once it reaches the second edge, it curves around the edge and reaches the third side. Let the rope be 3 3 times the side length of the hexagon.

Let the time taken to reach the first side be t 1 t_1 . Let the time to go from the first edge to the second edge be t 2 t_2 . Let the time to reach the third edge after reaching the second edge be t 3 t_3 . Let the total time of motion of the rope be T T .

Find ( t 2 ) ( T ) ( t 1 ) ( t 3 ) \frac{(t_2)(T)}{(t_1)(t_3)}


The answer is 6.

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2 solutions

Gaurang Pansare
Oct 22, 2014

Suppose the rope has initial angular velocity ' ω 1 \omega_1 '. After bending at 1st side, it becomes ' ω 2 \omega_2 ' and bending at 2nd side, it becomes ' ω 3 \omega_3 '

Let side of hexagon be ' a a '. Therefore length of rope = 3 a 3a

Now, the initial angle between rope and first face of hexagon is π 6 \frac{\pi}{6} . When rope completely covers one edge, it makes an angle of π 3 \frac{\pi}{3} with next edge. ω 1 = π 6 t 1 ω 2 = π 3 t 2 ω 3 = π 3 t 3 \therefore\;\;\;\; \omega_1=\frac{\frac{\pi}{6}}{t_1}\;\;\;\;\omega_2=\frac{\frac{\pi}{3}}{t_2}\;\;\;\;\omega_3=\frac{\frac{\pi}{3}}{t_3}

The linear velocity ' v v ' of the free end of the rope will be the same throughout as it is given so in the problem statement.

Now, v = 3 a ω 1 v = 2 a ω 2 v = a ω 3 t 1 = a π 2 v t 2 = 2 a π 3 v t 3 = a π 3 v \begin{matrix}v=3a*\omega_1&&v=2a*\omega_2&&v=a*\omega_3&&\\\therefore t_1=\frac{a\pi}{2v}&&\therefore t_2=\frac{2a\pi}{3v}&&\therefore t_3=\frac{a\pi}{3v}&&\end{matrix} T = t 1 + t 2 + t 3 = 3 a π 2 v \therefore T=t_1+t_2+t_3 = \frac{3a\pi}{2v} ( t 2 ) ( T ) ( t 1 ) ( t 3 ) = ( 2 a π 3 v ) ( 3 a π 2 v ) ( a π 2 v ) ( a π 3 v ) = 6 \therefore \frac{(t_2)(T)}{(t_1)(t_3)}= \frac{(\frac{2a\pi}{3v})(\frac{3a\pi}{2v})}{(\frac{a\pi}{2v})(\frac{a\pi}{3v})}= \boxed{6}

I have an alternative way of thinking about it.

I know that ω = 2 π f , ω = 2 π T , T = 2 π ω \omega =2\pi f\quad ,\quad \omega =\frac { 2\pi }{ T } ,\quad T=\frac { 2\pi }{ \omega }

I also know that v = ω r , ω = v r v=\omega r,\quad \omega =\frac { v }{ r }

Rearranging I get the following:

For T 1 = 2 π v r , T 1 = 2 π r v { T }_{ 1 }=\frac { 2\pi }{ \frac { v }{ r } } ,\quad { T }_{ 1 }=\frac { 2\pi r }{ v } .

As the motion progresses:

T 2 = 2 π 2 3 r v , T 3 = 2 π 1 3 r v , { T }_{ 2 }=\frac { 2\pi \cdot \frac { 2 }{ 3 } r }{ v } ,{ T }_{ 3 }=\frac { 2\pi \cdot \frac { 1 }{ 3 } r }{ v } ,

Here 'T' indicated the time period for a full circular path. However for T(1) only 1/12 of the period is completed and for T(2) and T(3) only 1/6 are completed. Hence

T 1 = π r 6 v , T 2 = 2 π r 9 v , T 3 = π r 9 v { T }_{ 1 }=\frac { \pi r }{ 6v } ,{ T }_{ 2 }=\frac { 2\pi r }{ 9v } ,{ T }_{ 3 }=\frac { \pi r }{ 9v }

T = 1.5 π r 9 v + 2 π r 9 v + π r 9 v = 9 π r 18 v = π r 2 v T=\frac { 1.5\pi r }{ 9v } +\frac { 2\pi r }{ 9v } +\frac { \pi r }{ 9v } =\frac { 9\pi r }{ 18v } =\frac { \pi r }{ 2v }

Following rearrangement into the form of the answer, and because we know that 'v' is constant and clearly the length of 'r' decreases by a third for each edge 'completed':

( T 2 ) ( T ) ( T 1 ) ( T 3 ) = 2 π r 9 v π r 2 v π r 6 v π r 9 v = 2 π 2 r 2 18 v 2 π 2 r 2 54 v 2 = 108 π 2 r 2 v 2 18 π 2 r 2 v 2 = 6 \frac { ({ T }_{ 2 })(T) }{ ({ T }_{ 1 }){ (T }_{ 3 }) } =\frac { \frac { 2\pi r }{ 9v } \cdot \frac { \pi r }{ 2v } }{ \frac { \pi r }{ 6v } \cdot \frac { \pi r }{ 9v } } =\frac { \frac { 2{ \pi }^{ 2 }{ r }^{ 2 } }{ 18{ v }^{ 2 } } }{ \frac { { \pi }^{ 2 }{ r }^{ 2 } }{ 54{ v }^{ 2 } } } =\frac { 108{ \pi }^{ 2 }{ r }^{ 2 }{ v }^{ 2 } }{ 18{ \pi }^{ 2 }{ r }^{ 2 }{ v }^{ 2 } } =6

In this case the solution is once again 6. :)

David Baker - 6 years, 7 months ago

Only one correction: the statement "The linear velocity '' of the free end of the rope will be the same throughout since no force is acting on it in that direction" doesn't seem right to me as this is not a dynamics problem. Linear velocity remains same because it is given so, not because of absense of some force. Nice solution otherwise. :)

Snehal Shekatkar - 6 years, 7 months ago

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point noted! changes made! :)

Gaurang Pansare - 6 years, 7 months ago

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u shud hv mentioned it is the linear velocity of free end

Mayank Shrivastava - 6 years, 6 months ago

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@Mayank Shrivastava Over rated problem!!!!!!!

Anubhav Tyagi - 5 years, 6 months ago
Austin Joseph
Dec 28, 2015

Just use conservation of angular momentum

W1(3r)^2 = W2(2r)^2=W3(r)^2

W=(theta/time)

9w1=4w2=w3. And now since theta is the same

9t3=t1 t2=4t3 T=14t3

now 56/9=6.2 or6

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