Question

Тело брошено с начальной скоростью 10 м/с под углом 60° к горизонту. Найдите нормальное и тангенциальное ускорения в начальный момент времени, а также радиус кривизны траектории в этот же момент.

Answer 1

Делаем рисунок траектории
Скорость направляем по касательной ( перпендикулярно радиусу)
ось Y направляем по скорости
ось X - по радиусу
рисуем вектор g
аn=g*соs60=10*0,5=5 м/с2
аt=g*sin60=10*√3/2=8,7 м/с2

аn=V^2/R

R=V^2/an=100/5=20 м

Answer 2

Calculation of Normal and Tangential Acceleration

To find the normal and tangential accelerations at the initial moment, we need to break down the initial velocity into its horizontal and vertical components.

Given: - Initial speed (v0) = 10 m/s - Launch angle (θ) = 60°

The horizontal component of the initial velocity (v0x) can be found using the formula:

v0x = v0 * cos(θ)

Substituting the given values:

v0x = 10 m/s * cos(60°) = 10 m/s * 0.5 = 5 m/s

The vertical component of the initial velocity (v0y) can be found using the formula:

v0y = v0 * sin(θ)

Substituting the given values:

v0y = 10 m/s * sin(60°) = 10 m/s * 0.866 = 8.66 m/s

Now, we can calculate the normal acceleration (an) and tangential acceleration (at) using the following formulas:

an = v^2 / R at = dv / dt

Where: - v is the magnitude of the velocity vector - R is the radius of curvature of the trajectory - dv is the change in velocity - dt is the change in time

At the initial moment, the velocity vector is tangent to the trajectory, so the normal acceleration is zero (an = 0).

The tangential acceleration (at) can be calculated using the formula:

at = dv / dt

Since the body is launched with an initial velocity and there is no change in velocity at the initial moment, the tangential acceleration is also zero (at = 0).

Therefore, the normal acceleration (an) and tangential acceleration (at) at the initial moment are both zero.

Calculation of Radius of Curvature

To find the radius of curvature (R) of the trajectory at the initial moment, we can use the formula:

R = v^2 / g

Where: - v is the magnitude of the velocity vector - g is the acceleration due to gravity (approximately 9.8 m/s^2)

The magnitude of the velocity vector (v) can be calculated using the Pythagorean theorem:

v = sqrt(v0x^2 + v0y^2)

Substituting the given values:

v = sqrt((5 m/s)^2 + (8.66 m/s)^2) = sqrt(25 m^2/s^2 + 75 m^2/s^2) = sqrt(100 m^2/s^2) = 10 m/s

Substituting the calculated value of v into the formula for the radius of curvature:

R = (10 m/s)^2 / 9.8 m/s^2 = 100 m^2/s^2 / 9.8 m/s^2 = 10.2 m

Therefore, the radius of curvature (R) of the trajectory at the initial moment is approximately 10.2 meters.

Note: The calculations assume that the body is moving in a vacuum without air resistance.