Question

Диск массой 20 кг и диаметром 0,4 м вращается со скоростью 120 об/мин и останавливается под действием момент трения 10 Нм. Величина момента импульса в конце второй секунды торможения

Answer 1

Ответ:

Объяснение:

Дано:

m = 20 кг

D = 0,4 м;      R = 0,2 м

n = 120 об/мин = 2 об/с

M = 10 Н·м

t = 2 c

________________

J·ω₂ - ?

1)

Момент инерции диска:

J = m·R²/2 = 20·0,2²/2 = 0,4 кг·м²

Угловая скорость:

ω₁ = 2·π·n ≈ 12,6 с⁻¹

2)

Основное уравнение динамики вращательного движения:

M·Δt = J·(ω₁ - ω₂)

Поскольку диск остановился, то ω₂ = 0

Δt = J·ω₁ / M = 0,4·12,6 / 10 = 0,5 c - время торможения меньше 2 секунд !

Answer 2

Calculation of Angular Momentum

To calculate the angular momentum at the end of the second second of braking, we need to use the formula for angular momentum:

Angular momentum (L) = Moment of inertia (I) × Angular velocity (ω)

The moment of inertia (I) depends on the mass (m) and the radius (r) of the rotating object. For a solid disk rotating about its central axis, the moment of inertia can be calculated using the formula:

Moment of inertia (I) = (1/2) × mass (m) × radius^2 (r^2)

Given: - Mass of the disk (m) = 20 kg - Diameter of the disk (d) = 0.4 m - Radius of the disk (r) = diameter/2 = 0.4/2 = 0.2 m - Initial angular velocity (ω) = 120 revolutions per minute (rpm) = 120 × 2π radians per minute - Time of braking (t) = 2 seconds - Frictional torque (τ) = 10 Nm

First, let's calculate the moment of inertia (I) using the given mass and radius:

I = (1/2) × m × r^2

Substituting the values:

I = (1/2) × 20 kg × (0.2 m)^2

Calculating the moment of inertia:

I = 0.4 kg·m^2 Next, let's convert the initial angular velocity from rpm to radians per second:

ω = 120 rpm × 2π radians/minute × (1/60) minutes/second

Calculating the angular velocity:

ω = 4π radians/second

Now, let's calculate the angular momentum (L) at the end of the second second of braking using the formula:

L = I × ω

Substituting the values:

L = 0.4 kg·m^2 × 4π radians/second

Calculating the angular momentum:

L = 1.6π kg·m^2/s

Therefore, the magnitude of the angular momentum at the end of the second second of braking is 1.6π kg·m^2/s.

Please note that the above calculations assume that the frictional torque remains constant throughout the braking process and that there are no external torques acting on the disk.