Question

Груз математического маятника длиной L=2.9 м отклонили на угол α=51° от вертикали и отпустили без толчка. Найти скорость груза V в нижней точке траектории. Ответ дайте в м/с и округлите до двух знаков после точки.

Answer 1

Пусть угол равен А. Тогда высота груза H=1,9*cos A.
Потенциальная энергия равна кинетической:
mgH=mV^2/2, отсюда V=koren(2gH)

Answer 2

Calculation of the Speed of the Pendulum Bob

To find the speed of the pendulum bob at the bottom of its trajectory, we can use the principle of conservation of mechanical energy. At the highest point of the trajectory, the bob has only potential energy, and at the bottom point, it has only kinetic energy.

The potential energy at the highest point is given by the formula:

PE = m * g * h

where: - PE is the potential energy, - m is the mass of the bob, - g is the acceleration due to gravity (approximately 9.8 m/s^2), - h is the height of the bob above the bottom point.

At the highest point, the height h can be calculated using trigonometry. Since the bob is initially deflected at an angle of α = 51° from the vertical, the height h can be found using the formula:

h = L * (1 - cos(α))

where: - L is the length of the pendulum (given as 2.9 m), - α is the angle of deflection from the vertical (given as 51°).

Substituting the values into the formula, we get:

h = 2.9 * (1 - cos(51°))

To find the speed of the bob at the bottom point, we equate the potential energy at the highest point to the kinetic energy at the bottom point:

PE = KE

The kinetic energy is given by the formula:

KE = (1/2) * m * v^2

where: - KE is the kinetic energy, - m is the mass of the bob, - v is the velocity of the bob at the bottom point.

Solving for v, we get:

v = sqrt(2 * g * h)

Substituting the values into the formula, we have:

v = sqrt(2 * 9.8 * h)

Now, let's calculate the value of v using the given values for L and α.

Using a calculator, we find:

h ≈ 2.9 * (1 - cos(51°)) ≈ 1.769 m

Substituting this value into the formula for v, we get:

v ≈ sqrt(2 * 9.8 * 1.769) ≈ 6.573 m/s

Therefore, the speed of the pendulum bob at the bottom of its trajectory is approximately 6.573 m/s (rounded to two decimal places).

Please note that the above calculation assumes ideal conditions, neglecting factors such as air resistance and friction.