Classical Mechanics 1 Name 142 Summative Assignment Sa 1

Classical Mechanics 1 Namema 142summative Assignment Sa 1numbersubmi

Consider the following problems related to vectors and motion in classical mechanics.

Instructions

Answer all questions clearly and thoroughly. Show all calculations and reasoning steps. Ensure your solutions are mathematically correct and presented in a logical manner.

Problem 1

Let i, j, k be the standard basis vectors. Consider the vectors a = 2i + 4j + 2k and b = i − j − 2k.

1. Calculate the dot product a · b and determine the angle α between vectors a and b. (Hint: Recall that cos^−1(−x) = π − cos^−1 x for any x.) [1 Mark]

2. Find a vector n that is orthogonal to both a and b and has a magnitude of 6. [1 Mark]

3. Given an origin fixed point C(1, 2, −2), find the distance from the origin to the plane p passing through C and parallel to both vectors a and b. (Hint: A vector orthogonal to the plane is orthogonal to two vectors parallel to the plane and not parallel to each other.) [1 Mark]

4. Let D = (1, −4, 1). Define plane q passing through the origin and parallel to p. Find the point E of intersection between the line CD and plane q. (Hint: Use the normal vector to q and the line equation.) [2 Marks]

Problem 2

A bee flies such that its velocity vector at time t ≥ 0 is v(t) = (6t + 1)i + (8t − 7)j − 3k. Let r(t) be the position vector of the bee, with initial position r(0) = 4i − 3j + 2k.

1. Determine r(t), the position vector of the bee. [1 Mark]

2. Find the time t > 0 when r(t) is orthogonal to the acceleration vector a(t). [1 Mark]

3. Identify the values of p in ℝ such that the vector b = r − tv + pt^2a remains constant with respect to t. [1 Mark]

4. At which time t > 0 does the acceleration a(t) have no tangential component? Additionally, compute the curvature of the trajectory at this time. (Hint: Compute |v(t)| and use the formula for the tangential component of acceleration.) [2 Marks]

Paper For Above instruction

Classical mechanics involves understanding vector operations, motion, and geometry in three-dimensional space. In addressing Problem 1, we analyze vector relationships, the geometry of planes, and their spatial relations. Problem 2 examines the dynamics of a moving particle, focusing on velocity, acceleration, and curvature of its trajectory.

Solution to Problem 1

Part 1: Dot Product and Angle between Vectors

The vectors are a = 2i + 4j + 2k, and b = i − j − 2k. The dot product a · b is given by:

a · b = (2)(1) + (4)(−1) + (2)(−2) = 2 - 4 - 4 = -6.

The magnitude of a is |a| = √(2^2 + 4^2 + 2^2) = √(4 + 16 + 4) = √24 = 2√6.

The magnitude of b is |b| = √(1^2 + (−1)^2 + (−2)^2) = √(1 + 1 + 4) = √6.

The cosine of the angle α between a and b is given by:

cos α = (a · b) / (|a| |b|) = -6 / (2√6 √6) = -6 / (2 6) = -6/12 = -0.5.

Therefore, α = cos^−1(−0.5) = 2π/3 radians or 120°.

Part 2: Finding a Vector Orthogonal to Both a and b with Magnitude 6

The vector orthogonal to both a and b is their cross product:

n = a × b

Calculating n:

n = |i j k|

|2 4 2|

|1 -1 -2|

Expanding along the first row:

n = i(4 (−2) - 2 (−1)) - j(2 (−2) - 2 1) + k(2 (−1) - 4 1)

n = i(−8 + 2) - j(−4 - 2) + k(−2 - 4)

n = i(−6) - j(−6) + k(−6) = (−6)i + 6j − 6k

To have magnitude 6, scale n:

|n| = √( (−6)^2 + 6^2 + (−6)^2 ) = √(36 + 36 + 36) = √108 = 6√3

Normalize to magnitude 6:

n' = (6 / 6√3) n = (1/√3) (−6i + 6j − 6k) = (−6/√3)i + (6/√3)j − (6/√3)k

which simplifies to:

n = (−2√3)i + 2√3j − 2√3k

Part 3: Distance from Origin to the Plane p

The plane p passes through point C(1, 2, −2) and is parallel to vectors a and b. The normal vector n to p is orthogonal to a and b, so n = (−6)i + 6j − 6k, as above.

The magnitude of n is |n| = 6√3, and a point on p is C.

The distance d from the origin to plane p is given by:

d = |(C · n̂)|, where n̂ is the unit normal vector:

n̂ = n / |n| = (−6i + 6j − 6k) / (6√3) = (−1/√3)i + (1/√3)j - (1/√3)k

Compute C · n̂:

C · n̂ = 1(−1/√3) + 2(1/√3) + (−2)*(−1/√3) = (−1/√3) + (2/√3) + (2/√3) = (−1 + 2 + 2)/√3 = 3/√3 = √3

Thus, the distance is:

d = |√3| = √3

Part 4: Intersection of Line CD and Plane q

Line CD passes through points C(1, 2, −2) and D(1, −4, 1). The vector from C to D is:

r_C D = D - C = (0, -6, 3)

The parametric form of line CD is:

r(t) = C + t(r_D - C) = (1, 2, -2) + t(0, -6, 3)

that is,

x = 1

y = 2 - 6t

z = -2 + 3t

Plane q passes through the origin with the same normal n as p: n = (−6, 6, −6).

The general equation of plane q is:

−6x + 6y − 6z = 0

Substituting the parametric coordinates:

−6(1) + 6(2 - 6t) − 6(-2 + 3t) = 0

-6 + 12 - 36t + 12 - 18t = 0

combining like terms:

(-6 + 12 + 12) + (-36t - 18t) = 0

18 - 54t = 0

t = 18 / 54 = 1/3

Now substitute into line equations:

x = 1

y = 2 - 6*(1/3) = 2 - 2 = 0

z = -2 + 3*(1/3) = -2 + 1 = -1

Therefore, E = (1, 0, -1).

Solution to Problem 2

Part 1: Position Vector r(t)

The velocity vector v(t) = (6t + 1)i + (8t − 7)j − 3k. To find r(t), integrate v(t) with respect to t:

r(t) = ∫ v(t) dt + r(0)

r(t) = ∫ (6t + 1) dt i + ∫ (8t - 7) dt j + ∫ (−3) dt k + r(0)

r(t) = (3t^2 + t)i + (4t^2 - 7t)j - 3t k + (4, -3, 2)

Part 2: Time when r(t) is orthogonal to a(t)

The acceleration vector a(t) is the derivative of v(t):

a(t) = d/dt v(t) = 6i + 8j + 0k

Compute the dot product r(t) · a(t):

r(t) · a(t) = [3t^2 + t, 4t^2 - 7t, -3t] · [6, 8, 0] = (3t^2 + t)6 + (4t^2 - 7t)8 + (-3t)*0

= 6(3t^2 + t) + 8(4t^2 - 7t) = 18t^2 + 6t + 32t^2 - 56t = (18t^2 + 32t^2) + (6t - 56t) = 50t^2 - 50t

Set equal to zero to find t:

50t^2 - 50t = 0

t(50t - 50) = 0

t = 0 or t = 1

Since t > 0, the relevant time is t = 1.

Part 3: Values of p for which b is constant

The vector b = r − t v + p t^2 a. Since r and v are known, and a is constant, analyze the dependence on t:

b(t) = r(t) - t v(t) + p t^2 a

Expressed explicitly:

r(t) = (3t^2 + t, 4t^2 - 7t, -3t)

t v(t) = t*(6t + 1, 8t - 7, -3) = (6t^2 + t, 8t^2 - 7t, -3t)

So, b(t) = (3t^2 + t - 6t^2 - t, 4t^2 - 7t - 8t^2 + 7t, -3t + 3t) + p t^2 a

which simplifies to:

b(t) = (−3t^2, -4t^2, 0) + p t^2 a

since a = (6, 8, 0), then p t^2 a = p t^2 (6, 8, 0) = (6p t^2, 8p t^2, 0)

Thus, b(t) = (−3t^2 + 6p t^2, -4t^2 + 8p t^2, 0)

For b to be independent of t, the coefficients of t^2 must be zero:

−3 + 6p = 0 → p = 1/2

−4 + 8p = 0 → p = 1/2

Therefore, p = 1/2 ensures b is constant in time.

Part 4: Time when acceleration has no tangential component and trajectory curvature

The tangential component of acceleration a_t = (a · v̂) v̂, where v̂ is the unit velocity vector. The magnitude of v(t):

|v(t)| = √{(6t + 1)^2 + (8t - 7)^2 + 9}

Compute this:

|v(t)|^2 = (6t + 1)^2 + (8t - 7)^2 + 9 = 36t^2 + 12t + 1 + 64t^2 - 112t + 49 + 9

= (36t^2 + 64t^2) + (12t - 112t) + (1 + 49 + 9) = 100t^2 - 100t + 59

Thus, |v(t)| = √(100t^2 - 100t + 59)

The acceleration vector a = (6, 8, 0) is constant, and its dot product with v(t) is:

a · v(t) = 6(6t + 1) + 8(8t − 7) = 36t + 6 + 64t - 56 = 100t - 50

The tangential component of acceleration is:

a_t = (a · v̂) v̂ = [ (a · v(t)) / |v(t)| ] * v̂

At the point when a · v̂ = 0, the tangential component vanishes, so:

100t - 50 = 0 → t = 0.5

Curvature κ at t = 0.5 can be calculated using:

κ = |v × a| / |v|^3

Calculate v × a:

v(t) = (6t + 1, 8t - 7, -3)

a = (6, 8, 0)

v × a = |i j k|

|6t+1 8t-7 -3|

|6 8 0|

Expanding:

i((8t - 7)0 - (-3)8) - j((6t + 1)0 - (-3)6) + k((6t + 1)8 - (8t - 7)6)

= i(0 + 24) - j(0 + 18) + k((48t + 8) - (48t - 42))

= 24i - 18j + (48t + 8 - 48t + 42)k = 24i - 18j + 50k

Magnitude |v × a| = √(24^2 + (−18)^2 + 50^2) = √(576 + 324 + 2500) = √3400 ≈ 58.31

|v| at t=0.5:

|v(0.5)| = √(100(0.5)^2 - 100(0.5) + 59) = √(25 - 50 + 59) = √34 ≈ 5.83

Finally, curvature:

κ = |v × a| / |v|^3 ≈ 58.31 / (5.83)^3 ≈ 58.31 / 198.25 ≈ 0.294

References

• Ampofo, B., & Sarpong, K. (2020). Classical mechanics for engineers. Springer.
• Fowles, G. R., & Cassiday, G. L. (2005). Analytical Mechanics. Thomson