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Student Neighbour Needs Help In Arriving At The Answer

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kuiperbelt | 08:17 Tue 25th Apr 2023 | Science
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Consider triangle OAB and let length of OA = a let angle ABO = A let length of AB = b and let angle AOB = B For this triangle the Sine Rule says that sin(A)/a = sin(B)/b ...... (1) Now, a = 0.5 m, A = φ, b = 1.5 m, B = 180° - θ and since sin(180° - θ) = sin(θ) equation (1) gives sin(φ)/0.5 = sin(θ)/1.5 Multiply both sides by 0.5: sin(φ) = 0.5 × sin(θ)/1.5 = sin(θ)/3 ...... (2) When...
12:07 Tue 25th Apr 2023
i dont get why your student neighbour doesn't ask his teachers how to do this stuff. Surely it's better to understand the method, and that is what he(if it's HE)/we(if it's FE) are paying for?
not that im miserable about it, i like looking at the q's. It just feels like this hel is already being paid for by someone but not being used
Getting tired of this person’s questions now.
Consider triangle OAB and
let length of OA = a
let angle ABO = A
let length of AB = b
and let angle AOB = B

For this triangle the Sine Rule says that
sin(A)/a = sin(B)/b ...... (1)

Now, a = 0.5 m, A = φ, b = 1.5 m, B = 180° - θ
and since sin(180° - θ) = sin(θ) equation (1) gives

sin(φ)/0.5 = sin(θ)/1.5

Multiply both sides by 0.5:

sin(φ) = 0.5 × sin(θ)/1.5 = sin(θ)/3 ...... (2)

When θ = 60°, sin(θ) = √3/2
therefore sin(φ) = √3/6
and φ = sin⁻¹(√3/6) ≈ 16.78°

It is a well-know property of functions of one independent variable that if f is a function of a variable u, and u is a function of a variable x, then
df/dx = df/du . du/dx
Differentiating both sides of (2) with respect to time, t, using this property gives

d(sin(φ))/dφ . dφ/dt = d(sin(θ)/3)/dθ . dθ/dt

For any x, d(sin(x))/dx = cos(x) so

cos(φ) . dφ/dt = (cos(θ)/3) . dθ/dt

Dividing both sides by cos(φ) gives

dφ/dt = (cos(θ)/3cos(φ)) . dθ/dt

When θ = 60°, cos(θ) = 0.5 and, as shown above, cos(φ) ≈ cos(16.78°) ≈ 0.9574
If dθ/dt = 10 rev/sec, multiply by 2π to convert to rad/sec giving
dθ/dt = 20π rad/sec and so

dφ/dt ≈ (0.5/(3 × 0.9574)) . 20π ≈ 10.94 rad/sec

QED
SMURF, if it's getting too much for you, don't look at any future questions.
PS - In my answer above I neglected to explain that the "rate at which the connecting rod is turning" is given by the rate of change of angle φ, expressed mathematically as dφ/dt, and "the crank makes 10 rev/sec" means that the rate of change of angle θ, expressed mathematically as dθ/dt, is 10 times 360° per second.
Don't they ever do any arts or humanities? Equally hard questions (if not harder) abound there, but I suppose students tend not to recognise that they are hard. :(
// I neglected to explain... dθ/dt, is 10 times 360° per second //

But you didn't, look...

dθ/dt = 10 rev/sec, multiply by 2π to convert to rad/sec giving

dθ/dt = 20π rad/sec

// "rate at which the connecting rod is turning" //

It's only right, the statement above is highlighted in inverted commas to dispel any ambiguity.

Since it is quite apparent that the connecting rod isn't turning (as in completing a full revolution), but is actually describing an arc.

One wonders if this might have been a misconception held by the student such as to cause his/her confusion?

If you ask me, BA is rightfully yours :-)





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Zebu>08.13 Fair Points!!

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