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Science Forum Index » Mathematics Forum » Angle of Wheel - Stumped...
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| Anon Email... |
 Posted: Thu Jun 12, 2008 6:54 pm |
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Guest
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There are two wheels, in a gear-like assemblage. One wheel is 5 cm
radius, the other is 10 cm radius. The large wheel is in a fixed
position, and cannot rotate. The small wheel sits on the large wheel,
at the very top of the wheel, and can traverse freely around the
circumference of the large wheel, rotating as it does so. The centre
of the small wheel is point A, and there is a line drawn straight down
from point A to point P on the circumference of the small wheel. The
centre of the large wheel is point B.
The small wheel starts its journey around the large wheel, and stops
when point A has travelled 8 cm from its origin (ie. an arc length).
What is the angle that the line AP now makes with the upward vertical?
Solution:
Let A1 be the original position of A
Let A2 be the resting place of point A
Let t be the angle A1 B A2
Let p the angle that the line AP now makes with the upward vertical
Arc length r*t = 8 cm (t = theta radians)
r = sum of both radii = 15 cm
Thus, t = 8/15 radians
Since the small circle rotates at twice the rate t increases, it will
have thus rotated 16/15 radians.
So the angle p is 16/15 - pi = .058 radians.
But my book gives the answer 1.542 radians. I have no idea why. Any
help greatly appreciated.
Cheers,
Anon |
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| Grover Hughes... |
| Posted: Fri Jun 13, 2008 1:46 pm |
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On Jun 12, 11:54 pm, Anon Email <anonema... at (no spam) fastmail.fm> wrote:
Quote: There are two wheels, in a gear-like assemblage. One wheel is 5 cm
radius, the other is 10 cm radius. The large wheel is in a fixed
position, and cannot rotate. The small wheel sits on the large wheel,
at the very top of the wheel, and can traverse freely around the
circumference of the large wheel, rotating as it does so. The centre
of the small wheel is point A, and there is a line drawn straight down
from point A to point P on the circumference of the small wheel. The
centre of the large wheel is point B.
The small wheel starts its journey around the large wheel, and stops
when point A has travelled 8 cm from its origin (ie. an arc length).
What is the angle that the line AP now makes with the upward vertical?
Solution:
Let A1 be the original position of A
Let A2 be the resting place of point A
Let t be the angle A1 B A2
Let p the angle that the line AP now makes with the upward vertical
Arc length r*t = 8 cm (t = theta radians)
r = sum of both radii = 15 cm
Thus, t = 8/15 radians
Since the small circle rotates at twice the rate t increases, it will
have thus rotated 16/15 radians.
So the angle p is 16/15 - pi = .058 radians.
But my book gives the answer 1.542 radians. I have no idea why. Any
help greatly appreciated.
Cheers,
Anon
The axle of wheel A is fastened to the axle of wheel B by what we can
call an "arm". Now imagine that all 3 objects (A, B, and the arm) are
locked together and given a rotation, say clockwise, of your 8/15
radian. At this point, each of the 3 parts has rotated 8/15 of a
radian clockwise. Now unlock the assembly, hold the arm fixed, and
give wheel B a counter-clockwise rotation of 8/15 radian, which
returns it to its original position. During this 8/15 radian rotation,
wheel A must turn an additional amount of twice 8/15, i.e., 16/15
radian, again clockwise. The total rotation of wheel A is then the sum
of the original 8/15 plus the final 16/15, or 24/15 radian. This
amount (24/15, or 1.6 radian) doesn't agree with your book, so
possibly I have misunderstood your description of the setup. Or,
possibly, the book is wrong ......?
HTH,
Grover Hughes |
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| Greg Neill... |
| Posted: Fri Jun 13, 2008 10:48 pm |
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"Grover Hughes" <ghughes at (no spam) cei.net> wrote in message
news:7558608b-78af-4357-a8c3-08054dece125 at (no spam) f63g2000hsf.googlegroups.com
Quote: On Jun 12, 11:54 pm, Anon Email <anonema... at (no spam) fastmail.fm> wrote:
There are two wheels, in a gear-like assemblage. One wheel is 5 cm
radius, the other is 10 cm radius. The large wheel is in a fixed
position, and cannot rotate. The small wheel sits on the large wheel,
at the very top of the wheel, and can traverse freely around the
circumference of the large wheel, rotating as it does so. The centre
of the small wheel is point A, and there is a line drawn straight
down from point A to point P on the circumference of the small
wheel. The centre of the large wheel is point B.
The small wheel starts its journey around the large wheel, and stops
when point A has travelled 8 cm from its origin (ie. an arc length).
What is the angle that the line AP now makes with the upward
vertical?
Solution:
Let A1 be the original position of A
Let A2 be the resting place of point A
Let t be the angle A1 B A2
Let p the angle that the line AP now makes with the upward vertical
Arc length r*t = 8 cm (t = theta radians)
r = sum of both radii = 15 cm
Thus, t = 8/15 radians
Since the small circle rotates at twice the rate t increases, it will
have thus rotated 16/15 radians.
So the angle p is 16/15 - pi = .058 radians.
But my book gives the answer 1.542 radians. I have no idea why. Any
help greatly appreciated.
Cheers,
Anon
The axle of wheel A is fastened to the axle of wheel B by what we can
call an "arm". Now imagine that all 3 objects (A, B, and the arm) are
locked together and given a rotation, say clockwise, of your 8/15
radian. At this point, each of the 3 parts has rotated 8/15 of a
radian clockwise. Now unlock the assembly, hold the arm fixed, and
give wheel B a counter-clockwise rotation of 8/15 radian, which
returns it to its original position. During this 8/15 radian rotation,
wheel A must turn an additional amount of twice 8/15, i.e., 16/15
radian, again clockwise. The total rotation of wheel A is then the sum
of the original 8/15 plus the final 16/15, or 24/15 radian. This
amount (24/15, or 1.6 radian) doesn't agree with your book, so
possibly I have misunderstood your description of the setup. Or,
possibly, the book is wrong ......?
HTH,
Grover Hughes
The answer is indeed 1.542 radians.
Say that the center of the smaller circle travels from
its initial position A to A', an arc distance s1 (8cm).
The angle of the displacement of A to A' as observed
from the center of the larger circle, B, is then
t = s1/(r1 + r2)
where r1 is the radius of the larger circle, 10cm
r2 is the radius of the smaller circle, 5cm
The smaller circle is assumed to remain in contact
with the larger circle as it moves (it rolls). The
arc length traversed on the larger circle by the
contact point is thus
s2 = r1*t
The smaller circle, in contact with the larger
circle, will have rolled this same distance along
its circumference, carrying the initial point of
contact p to position p' on the smaller circle,
an arclength of s2 on the smaller circle displaced
from the new point of contact between the two
circles.
An arclength of s2 on the smaller circle
corresponds to an angle
a = s2/r2
for the ankle BA'p'. That's the angle at the new center
of the smaller circle, A', from the center of the
larger circle to the new point p'.
If the line A'p' is extended to the vertical of the
line BA (the vertical line passing through the initial
centers of the circles), and we call the intersection C,
triangle BA'C is formed. Call the angle BCA' b. We
have then
t + a + b = pi
To solve the problem posed we want to know angle b.
Thus
b = pi - a - t
= pi - s2/r2 - s1/(r1 + r2)
Simplifying:
b = (pi*r2 - s1)/r2 = ~1.542 |
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| Anon Email... |
| Posted: Sat Jun 14, 2008 4:20 pm |
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Guest
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Thanks guys. I'll look at this in more detail.
Cheers,
Anon |
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