Is this possible? (1 Viewer)

[NJMORTIMER]

I got this question from my ext. 2 book and am having major problems seeing how to solve it.

Q: "If the number of revolutions per minute of a conical pendulum is increased from 75 to 80, show that the rise in the level of the bob is approximately 19.2mm"

It seems like some piece of information is missing (i.e. the string length), otherwise what would happen if the string length was less than 19.2 mm?!
I tried taking the result of a rise of 19.2mm and reverse engineering the question to find the string length that would result in this rise but alas all I managed to prove is that 0.0192=0.3999 :rofl:

Any insight/help would be muchly appreciated!

 

[Trebla]

The length of the string is irrelevent. In the case of the string being less 19.2mm, it would rise to form an 'upside down' conical pendulum (i.e. beyond the horizontal).
Let:
θ = angle between string and vertical
T = tension of string
r = radius of circular motion
m = mass of bob
g = acceleration due to gravity
ω = angular velocity
h = vertical height from one end of the string to the bob

By resolving the forces:
Horizontal component:
Tsin θ = mrω² (1)
Vertical component:
Tcos θ = mg (2)
(1)/(2): tan θ = (rω²)/g
but tan θ also equals r/h using the right-angled triangle enclosed by h, r and the string
r/h = (rω²)/g
h = g/ω²

Now simply convert 75 revolutions per minute into radians per second and sub it into h (call this h1). Repeat with 80 revolutions per minute (call this h2). Find their difference (h1 - h2), and it should equal 19.2mm

You should get:
h1 ≈ 0.15887... m
≈ 158.87 mm
h2 ≈ 0.13963... m
≈ 139.63mm
h1 - h2 ≈19.24 mm ≈ 19.2 mm

 

[NJMORTIMER]

Wow thanks heaps for that, good explanation, but there are a few things that don't make sense to me. How is it possible to have a weighted conical pendulum rise above horizontal (I can imagine a sudden increase in rev/min would 'swing' it above the horizontal but why would it stay there)? Also, what if the string was 5mm long, even going from straight down to straight up would now allow for a difference of 18.2mm. Lastly, the units for h = g/ω², suggests the height 'h' is measured in metres per square radian 'g/ω²'?
I'm finding this really interesting and really confusing at the same time
Again, your explanation is greatly appreciated but please help me to understand the rest.

 

[Trebla]

Wow thanks heaps for that, good explanation, but there are a few things that don't make sense to me. How is it possible to have a weighted conical pendulum rise above horizontal (I can imagine a sudden increase in rev/min would 'swing' it above the horizontal but why would it stay there)? Also, what if the string was 5mm long, even going from straight down to straight up would now allow for a difference of 18.2mm.

The string will stay there as long as the tension/centripetal force allows it to do so. Have you ever tried spinning a ball attached to a string really fast above your head? It looks just like that.
Obviously with 5mm it wouldn't work. The value of h is restricted by the length of the string. However, in the context of exam problems, I don't think that you'll ever face such a scenario.

Lastly, the units for h = g/ω², suggests the height 'h' is measured in metres per square radian 'g/ω²'?
I'm finding this really interesting and really confusing at the same time

Yes, BUT remember that radians refers to the arc displacement in circular motion only and the metre unit does not. Using the fact that v = rω to relate radians and metres, substitute for ω and you should get h as a unit in metres.