Question

Материальная точка движется по окружности радиусом r = 2 м согласно уравнению φ = at+bt3, где a = 8 рад/с, b = –0.2 рад/с3. Найти тангенциальное ускорение в момент времени t = 3 с. Ответ выразить в СИ.

Answer 1

Подставляем значения a и b в уравнение:
φ = at + bt³ = 8t - 0,2t³.
Угловая скорость равна ω = dφ/dt = 8 - 0,2*3t² = 8 - 0,6t².
Угловое ускорение равно ε = dω/dt = - 0,6*2t = -1,2t.
Тангенциальное ускорение равно aτ = ε*r = -1,2t*r = -1,2*3 c *2 м = -7,2 м/с².

Answer 2

Calculation of Tangential Acceleration

To find the tangential acceleration of a material point moving along a circular path, we need to differentiate the equation φ = at + bt^3 with respect to time (t) twice. The first derivative will give us the tangential velocity, and the second derivative will give us the tangential acceleration.

Given: - Radius of the circular path (r) = 2 m - Equation of motion: φ = at + bt^3 - Constants: a = 8 rad/s, b = -0.2 rad/s^3 - Time (t) = 3 s

Let's calculate the tangential acceleration.

Derivation of Tangential Velocity

Differentiating the equation φ = at + bt^3 with respect to time (t), we get:

dφ/dt = a + 3bt^2

This represents the tangential velocity (v) of the material point.

Derivation of Tangential Acceleration

Differentiating the equation dφ/dt = a + 3bt^2 with respect to time (t), we get:

d^2φ/dt^2 = 6bt

This represents the tangential acceleration (a_t) of the material point.

Calculation of Tangential Acceleration at t = 3 s

Substituting the given time (t = 3 s) into the equation d^2φ/dt^2 = 6bt, we can calculate the tangential acceleration.

Given: - Time (t) = 3 s - Constant (b) = -0.2 rad/s^3

Substituting the values into the equation, we have:

a_t = 6b * t = 6 * (-0.2 rad/s^3) * 3 s = -0.36 rad/s^2

Therefore, the tangential acceleration at t = 3 s is -0.36 rad/s^2.

Note: The negative sign indicates that the tangential acceleration is directed opposite to the direction of motion.

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