## Description

Need help with the files I uploaded for homework.

Would like if possible the best formula to work out the problems and step by step how you solved them.

Q1 and Q2 are related to the Intro file.

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## Explanation & Answer

I added the solution for those parameters. In the process I found a mistake (previously I chose the vector u in opposite direction to the desired).

0. The terms.

̂ . Also

Denote the plane of rotation 𝑅, the center of rotation 𝐶, the unit normal vector to 𝑅 at 𝐶 as 𝒏

̂ and parallel to the 𝑥𝑦-plane as 𝒖

̂ . The angle between the

denote the unit vector perpendicular to 𝒏

radius-vector 𝒓 will be called 𝜃 and the angle corresponding to time 𝑡 = 0 will be 𝜃0 . Let the radius

of rotation be 𝑟, the (constant) angular speed be 𝜔.

The position of 𝑅 relative to the world frame of reference.

̂ makes with the 𝑥𝑦-plane and the angle 𝜓 that the

It may be specified by the angle 𝜑 that 𝒏

̂ to the 𝑥𝑦-plane makes with the 𝑥 (to the right) axis.

projection of 𝒏

1.

a. Denote the coordinates of the point 𝐶 as (0, 0, ℎ) in the standard world frame of reference. It is

given that 1 ≤ ℎ ≤ 1.5 (in meters).

̂ on the 𝑧-axis is sin 𝜑, the projection on the 𝑥𝑦-plane has the length cos 𝜑

b. The projection of 𝒏

hence the 𝑥-projection is cos 𝜑 cos 𝜓 and the 𝑦-projection is cos 𝜑 sin 𝜓.

̂ = cos 𝜑 cos 𝜓 𝒊 + cos 𝜑 sin 𝜓 𝒋 + sin 𝜑 𝒌.

𝒏

̂ lies in 𝑅 (parallel to it). If 𝑅 is not horizontal (which I believe is

c. Any vector that is orthogonal to 𝒏

̂.

true), then there is only one unit vector in it which is parallel to the ground (𝑥𝑦-plane). Denote it 𝒖

̂ = 𝑎𝒊 + 𝑏𝒋 and 𝒖

̂⊥𝒏

̂ , i.e. 𝑎 cos 𝜑 cos 𝜓 + 𝑏 cos 𝜑 sin 𝜓 = 0 or

We know that 𝒖

̂ = − sin 𝜓 𝒊 + cos 𝜓 𝒋.

𝑎 cos 𝜓 + 𝑏 sin 𝜓 = 0. Because we need a unit vector, 𝒖

̂ and 𝒖

̂ , use cross product 𝒖

̂⊥ = 𝒏

̂×𝒖

̂.

d. To obtain a third vector orthogonal to both 𝒏

̂.

It lies in 𝑅 because it is orthogonal to 𝒏

𝒊

̂⊥ = 𝒏

̂×𝒖

̂ = |cos 𝜑 cos 𝜓

𝒖

− sin 𝜓

𝒋

cos 𝜑 sin 𝜓

cos 𝜓

𝒌

sin 𝜑| = − sin 𝜑 cos 𝜓 𝒊 − sin 𝜑 sin 𝜓 𝒋 + cos 𝜑 𝒌.

0

̂ and 𝒖

̂ ⊥ make an orthonormal basis in 𝑅.

This way the vectors 𝒖

2.

̂ , as 𝜃0 . The

a. Denote the angle corresponding to 𝑡 = 0, which the rope makes with the vector 𝒖

angular speed 𝜔 in radians per second is supposed to be constant.

Then the angular displacement is linear: 𝜃(𝑡) = 𝜃0 + 𝜔𝑡.

̂, 𝒖

̂ ⊥ of a vector 𝒗

̂ in 𝑅 that makes angle 𝜃 with the vector 𝒖

̂ are obviously

b. The projections on 𝒖

𝑟 cos 𝜃 and 𝑟 sin 𝜃. This means

̂ = 𝑟 cos 𝜃 ∙ 𝒖

̂ + 𝑟 sin 𝜃 ∙ 𝒖

̂⊥.

𝒗

c. In terms of time 𝑡 we obtain

̂ + 𝑟 sin(𝜃0 + 𝜔𝑡) ∙ 𝒖

̂⊥.

𝑃(𝑡) = 𝑟 cos(𝜃0 + 𝜔𝑡) ∙ 𝒖

d. Almost the same:

̂ + 𝑟 cos(𝜃0 + 𝜔𝑡) ∙ 𝒖

̂ + 𝑟 sin(𝜃0 + 𝜔𝑡) ∙ 𝒖

̂⊥.

𝒓( 𝑡 ) = 0 ∙ 𝒏

Or we can write it as

0

𝒓(𝑡) = (𝑟 cos(𝜃0 + 𝜔𝑡))

𝑟 sin(𝜃0 + 𝜔𝑡)

̂, 𝒖

̂, 𝒖

̂ ⊥ }.

with respect to the basis {𝒏

3.

̂, 𝒖

̂, 𝒖

̂ ⊥ } in terms of {𝒊, 𝒋, 𝒌}. The required matrix consists of these

a. We know how to express {𝒏

decompositions as columns:

cos 𝜑 cos 𝜓

𝐴 = ( cos 𝜑 sin 𝜓

sin 𝜑

− sin 𝜓

cos 𝜓

0

− sin 𝜑 cos 𝜓

− sin 𝜑 sin 𝜓 ).

cos 𝜑

̂.

b. The set of equations is obviously 𝒊 = 𝒓̂, 𝒋 = 𝒇̂, 𝒌 = −𝒅

0 1

Hence the transformation matrix is 𝐵 = (1 0

0 0

0

0 ).

−1

c. Well, let the camera be 𝑚 meters behind the point of rotation: (0, −𝑚, ℎ) in world frame.

The vector to 𝐶 is (0, 𝑚, 0).

d. To convert 𝒓(𝑡) known in 𝑅 coordinates we need to perform the transformation given by 𝐴 and

then by 𝐵:

− sin 𝜓 𝑟 cos(𝜃0 + 𝜔𝑡) − sin 𝜑 cos 𝜓 𝑟 sin(𝜃0 + 𝜔𝑡)

0

(𝑟 cos(𝜃0 + 𝜔𝑡)) ↦ ( cos 𝜓 𝑟 cos(𝜃0 + 𝜔𝑡) − sin 𝜑 sin 𝜓 �...