Curvilinear Motion

Question 1

What is the definition of trajectory in the context of motion?

Answer 1

The path traced by a moving object, outlining the specific route it follows through space.

Question 2

What does the radius of curvature represent in motion?

Answer 2

It represents the radius of the best-fitting circular arc to a curve at a specific point.

It indicates how sharply a path bends, with a smaller radius corresponding to a sharper turn and a larger radius indicating a gentler curve.

Mathematically, the radius of curvature (𝑅) is related to the rate of change of the direction of motion and can be calculated using the formula:

𝑅 = 𝑣² / |𝑎ₙ|

where:
𝑣 is the speed of the object,
𝑎ₙ (normal acceleration) is the component of acceleration perpendicular to the velocity, responsible for changing the direction of motion.

Question 3

What does the tangential component of acceleration represent?

Answer 3

The change in speed along the direction of motion.

It determines how fast or slow an object moves along its trajectory.

Question 4

What does the normal component of acceleration describe?

Answer 4

The change in direction of motion.

It acts perpendicular to the velocity and is responsible for curved motion.

Question 5

What advantage do Cartesian coordinate system provide when analyzing motion in two dimensions?

Answer 5

It allows motion to be studied by treating each direction independently, simplifying the analysis of complex trajectories by resolving motion into separate 𝑥- and 𝑦- components.

Question 6

Solve the problem:

A car is moving along a curved road with a speed of 30 m/s. If the tangential acceleration is 4 m/s² and the radius of curvature is 50 meters, what is the normal acceleration of the car?

Answer 6

18 m/s²

We can use the formula for normal acceleration:

𝑎ₙ = 𝑣² / 𝑟

Substituting the given values:

𝑎ₙ = (30 𝑚/𝑠)² / 50 𝑚 = 900 / 50 = 18 𝑚/𝑠²

The normal acceleration is 18 m/s².

Question 7

Solve the problem:

A projectile is launched with an initial velocity of 20 m/s at an angle of 45°. What is the horizontal range of the projectile?

Answer 7

40.82 meters

The horizontal range 𝑅 of a projectile is given by the formula:

𝑅 = (𝑣₀² sin(2𝜃)) / 𝑎

where:

𝑣₀ = 20m/s
𝜃 = 45°
𝑎 = 9.8m/s²

𝑅 = 400 / 9.8 = 40.82

The horizontal range is 40.82 meters.

Question 8

Solve the problem:

An object moves along a curvilinear path. The velocity components in the 𝑥 and 𝑦- directions are 𝑣ₓ = 4𝑥 and 𝑣ᵧ = 6𝑥, respectively. Determine the total velocity of the object at the point where 𝑥 = 2.

Answer 8

14.42 m/s

To find the total velocity, we need to calculate the magnitude of the velocity vector.

Find the velocity components at 𝑥 = 2.

Given the velocity components:
𝑣ₓ = 4𝑥
𝑣ᵧ = 6𝑥

At 𝑥 = 2, the velocity components are:
𝑣ₓ = 4(2) = 8 𝑚/𝑠
𝑣ᵧ = 6(2) = 12 𝑚/𝑠

The total velocity 𝑣 is the magnitude of the velocity vector:
𝑣 = √(𝑣ₓ² + 𝑣ᵧ²)

Substitute the values:
𝑣 = √(8² + 12²) = √(64 + 144) = √208 ≈ 14.42 𝑚/𝑠

The total velocity of the object at the point where
𝑥 = 2 is approximately 14.42 m/s.

Question 9

Solve the problem:

A car is moving along a curved path with an initial speed of 25 m/s. The car experiences a tangential acceleration of 3 m/s². How much speed will the car have after 6 seconds?

Answer 9

43m/s

We use the equation for velocity under constant acceleration:

𝑣 = 𝑣₀ + 𝑎ₜ ⋅ 𝑡

Substitute the values:

𝑣 = 25 + 3 ⋅ 6
= 25 + 18
= 43 m/s

The car will have a speed of 43m/s after 6 seconds.

Question 10

Solve the problem:

A particle has a tangential acceleration of 3 m/s² and a normal acceleration of 7 m/s². What is the magnitude of the total acceleration?

Answer 10

7.62 m/s²

The total acceleration 𝑎 is the vector sum of the tangential and normal accelerations. Using the Pythagorean theorem:

𝑎 = √(𝑎ₜ² + 𝑎ₙ²)

Substitute the values:

𝑎 = √(3² + 7²)
= √(9 + 49)
= √58
≈ 7.62 m/s²

The total acceleration is approximately 7.62 m/s².