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प्रश्न
Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in cartesian co-ordinates A = `A_xhati + A_yhatj` where `hati` and `hatj` are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of (Figure). Motion can also be studied by expressing vectors in circular polar co-ordinates as A = `A_rhatr + A_θhatθ` where `hatr = r/r = cos θhati + sin θj` and `hatθ = - sin θhati + cos θ hatj` are unit vectors along direction in which `r` and `θ` are increasing.
- Express `hati` and `hatj` in terms of `hatr` and `hatθ`
- Show that both `hatr` and `hatθ` are unit vectors and are perpendicular to each other.
- Show that `d/(dt) (hatr) = ωhatθ` where `θ = (dθ)/(dt)` and `d/(dt) (hatθ) = - ωhatr`
- For a particle moving along a spiral given by `t = aθhatr`, where a = 1 (unit), find dimensions of ‘a’.
- Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.
उत्तर
a. Given, unit vetor `hatr = cos θhati + sinθhatj` ......(i)
`hatθ = - sin θ hati + cos θhatj` ......(ii)
Multiplying equation (i) by sin θ and equation (ii) with cos θ and adding
`hatr sin θ + hatθ cos θ = sin θ. cos θhati + sin^2 θhatj + cos^2 θhatj - sinθ. cos θhati`
= `hatj (cos^2 θ + sin^2 θ) = hatj`
⇒ `hatr sin θ + θ cos θ = hatj`
By equation (i) `xx cos θ` - equation (ii) `xx sin θ`
`n(hatr cos θ - hatθ sin θ) = hati`
b. `hatr.hatθ = (cos θhati + sinθhatj) . (- sin θhati + cosθhatj)`
= `- cos θ . sin θ + sin θ . cos θ`
= 0
⇒ θ = 90° Angle between `hatr` and `hatθ`
c. Given, `hatr = cos θhati + sinθhatj`
`(dhatr)/(dt) = d/(dt) (cos θhati + sinθhatj)`
= `- sin θ . (dθ)/(dt) hati + cos θ . (dθ)/(dt) hatj`
= `ω [- sin θhati + cos θhatj]` ......`[∵ θ = (dθ)/(dt)]`
d. Given, `r = aθhatr`, here, writing dimensions `[r] = [a][θ][hatr]`
⇒ L = [a] % 1
⇒ L = [M0L1T0]
e. Given, a = 1 unit `r = θhatr = θ[cos θhati + sin θhatj]`
Velocity, v = `(dr)/(dt) = (dθ)/(dt) hatr + θ d/(dt) hatr = (dθ)/(dt) hatr + θ d/(dt) [(cos θhati + sinθhatj)]`
= `(dθ)/(dt) hatr + θ [(- sin hati + cos θ hatj) (dθ)/(dt)]`
= `(dθ)/(dt) hatr + θ hatθ ω`
= `ωhatr + ω θhatθ`
Acceleration, a = `d/(dt) [ωhatr + ωθhatθ]`
= `d/(dt) [(dθ)/(dt) hatr + (dθ)/(dt) (θhatθ)]`
= `(d^2θ)/(dt^2) hatr + (dθ)/(dt) * (dhatr)/(dt) + (d^2θ)/(dt^2) θhatθ + (dθ)/(dt) d/(dt) (θhatθ)`
= `(d^2θ)/(dt^2) hatr + ω[-sin θhati + sinθhatj] + (d^2θ)/(dt^2) θhatθ + (ωd)/(dt) (θhatθ)`
= `(d^2θ)/(dt^2) hatr + ω^2hatθ + (d^2θ)/(dt^2) xx θhatθ + ω^2hatθ + w^2θ (-hatr) ((d^2θ)/(dtv^2) - ω^2)hatr + (2ω^2 + (d^2θ)/(dt^2) θ)θ`
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