Solution 20.2.2 from the collection (workbook) of Kepe O.E. 1989

Solution tasks 20.2.2

Consider a uniform rod of length l = 3 m and mass m = 30 kg, rotating in a vertical plane. It is necessary to find the generalized force corresponding to the generalized coordinate Phi, at the moment when the angle Phi = 45°.

Generalized force corresponding to a generalized coordinate Phi, is determined by the formula:

Q = p_Phi - d/dt(∂L/∂Phi),

Where p_Phi - generalized impulse, L - Lagrangian, t - time.

To find the generalized momentum, we use the formula:

p_Phi = ∂L/∂(dPhi/dt).

To find the Lagrangian, we write down the kinetic and potential energies of the system:

T = (ml²/3)(dPhi/dt)²,

U = 0.

Then the Lagrangian will have the form:

L = T - U = (ml²/3)(dPhi/dt)².

Let us differentiate the Lagrangian with respect to time:

d/dt(∂L/∂(dPhi/dt)) - ∂L/∂Phi = 0.

Let's substitute the values L and Phi:

(ml²/3)·2(d²Phi/dt²) - 0 = 0.

Where do we get it from:

(d²Phi/dt²) = 0.

Thus, the generalized impulse will be equal to:

p_Phi = ∂L/∂(dPhi/dt) = 2(ml²/3)(dPhi/dt).

Now we find the derivative of the Lagrangian with respect to time:

d/dt(∂L/∂(dφ/dt)) = (ml²/3)·2(d³ph/dt³).

Substituting values p_φ and d/dt(∂L/∂φ) into the formula for the generalized force:

Q = p_φ - d/dt(∂L/∂φ) = 2(ml²/3)(dφ/dt) - 0 = 2(ml²/3)(dφ/dt).

Let's find the value of the angle φ at which this force will be maximum, that is, when the derivative dQ/dφ will be equal to zero:

dQ/dφ = 2(ml²/3)·(d²φ/dt²) = 0.

Whence it follows that d²φ/dt² = 0, that is, the angle φ will be permanent. Thus, the maximum force will be achieved at any time at an angle φ = 45°. The force value will be equal to:

Q = 2(ml²/3)(dφ/dt) = 2·(30 kg)·(3 m)²/(3·2)·(Pi/4 rad/s) ≈ 706 N.

Thus, when a homogeneous rod 3 m long and weighing 30 kg rotates in a vertical plane, the generalized force corresponding to the generalized coordinate of the angle φ, at the moment when the angle φ equal to 45°, will be equal to approximately 706 N.

Product description: Solution 20.2.2 from the collection (resolution book) by Kepe O.E. 1989

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Solution 20.2.2 from the collection (workbook) of Kepe O.E. 1989 presents a solution to the problem of system dynamics with one degree of freedom. The problem considers a homogeneous rod 3 meters long and weighing 30 kg, which rotates in a vertical plane. It is required to determine the generalized force corresponding to the generalized coordinate φ at the moment of time when the angle φ is equal to 45°.

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