A centripetal-acceleration addict rides in uniform circular motion with period T = 4.30 s and radius r = 3.00 m. At one instant his acceleration is a = (4.00 m/s2)i + (-5.00 m/s2)j . At that instant, what are the following values?

Question

A centripetal-acceleration addict rides in uniform circular motion with period T = 4.30 s and radius r = 3.00 m. At one instant his acceleration is a = (4.00 m/s2)i + (-5.00 m/s2)j . At that instant, what are the following values?

(a) v · a

(b) r x a

Answer 1

To find the values of (a) v · a and (b) r x a at the given instant, we need to follow these steps:

Step 1: Find the velocity vector v:
Since the motion is uniform circular motion, we can use the formula for the velocity of an object in circular motion: v = ωr, where ω is the angular velocity and r is the radius.

To find the angular velocity ω, we can use the formula ω = 2π/T, where T is the period.

Given that T = 4.30 s, we can calculate ω as follows:
ω = 2π / T = 2π / 4.30 s ≈ 1.463 rad/s

Now we can calculate the velocity v using the formula v = ωr:
v = ωr = (1.463 rad/s)(3.00 m) ≈ 4.389 m/s

So we have the velocity vector v = (4.389 m/s, 0 m/s)

Step 2: Find the values of (a) v · a:
To find the dot product of two vectors, we multiply their corresponding components and then sum them up.

Given the acceleration vector a = (4.00 m/s², -5.00 m/s²), we can calculate (a) v · a as follows:
v · a = (4.389 m/s)(4.00 m/s²) + (0 m/s)(-5.00 m/s²) = 17.556 m/s²

So the value of (a) v · a is 17.556 m/s².

Step 3: Find the value of (b) r x a:
To find the cross product of two vectors, we use the formula r x a = |r| |a| sin(θ) n, where |r| and |a| are the magnitudes of the vectors, θ is the angle between them, and n is a unit vector perpendicular to both r and a.

Given |r| = 3.00 m and |a| = √(4.00 m/s²)² + (-5.00 m/s²)² ≈ 6.403 m/s², we need to find the angle θ between r and a.

To find θ, we can use the formula cos(θ) = a · r / (|a| |r|):
cos(θ) = (4.00 m/s²)(3.00 m) / (6.403 m/s²)(3.00 m) ≈ 0.371
θ ≈ arccos(0.371) ≈ 67.389 degrees

Now we can calculate (b) r x a using the formula r x a = |r| |a| sin(θ) n:
r x a = (3.00 m)(6.403 m/s²) sin(67.389 degrees) n

Note: We can only determine the magnitude and direction of the cross product, not the explicit vector values. Thus, the answer will be in the form of a magnitude and a direction vector.

So the value of (b) r x a is (approximately) 19.21 m²/s² n.