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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
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
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
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.
// 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 :-)
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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