A long uniform rod of length 6.00 m and mass 6.00 kg is pivoted about a horizont

Question

A long uniform rod of length6.00m and mass6.00kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.61.

A long uniform rod of length6.00m and mass6.00kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.61. Figure P10.61 (a) What is the angular speed of the rod at the instant it is horizontal? Correct: Your answer is correct. (b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal? Correct: Your answer is correct. (c) Find the components of the acceleration of the rod's center of mass. ax=Correct: Your answer is correct. ay=Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. (d) Find the components of the reaction force at the pivot. Rx=Correct: Your answer is correct. Ry=Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 100%. I only need help on finding a(y) and r(y). The solutions for a(x) and r(x) are correct as shown. Please show work and round to four decimal places if possible.

Explanation / Answer

Weight of rod acts from its centre. Let denote the angle from vertical as the rod falls.

Torque = mg*L/2*Sin = 4*9.81*6/2*Sin = 117.72*Sin N-m

Inertia of rod I = mL2/3 = 4*62/3 = 48 kg-m2

Angular acceleration = Torque/Inertia = 117.72*Sin/48 = 2.4525*Sin rad/s2

(a) = d/dt = 2.4525*Sin

so, [d/d](d/dt) = 2.4525*Sin

so, (d/d) = 2.4525*Sin

Integrating on both the sides, 2/2 = -2.4525*Cos + C

Since = 0 at = 0, we get, C = 2.4525

So, 2/2 = -2.4525*Cos + 2.4525

When rod is horizontal, = /2, so 2/2 = -2.4525*Cos(/2) + 2.4525

So, = 2.215 rad/s

(b) = d/dt = 2.4525*Sin

when rod is horizontal, = /2, so = 2.4525*sin(/2) = 2.4525 rad/s2

(c) Ay = -(L/2)* = -(6/2)*2.4525 = -7.3575 m/s2

Ax = (L/2)*2 = (6/2)*2.2152 = 14.7 m/s2

(d) Ry = -m*Ay - mg = -4*7.3575 - 4*9.81 = -68.67 N

Rx = m*Ax = 4*14.7 = 58.8 N

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