A thin rod (uniform density & thickness) has mass M and length L. It is attached to the floor at a fixed location by a friction-less hinge. The rod start at rest, balanced vertically on its hinge end.
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1) While balanced vertically, the gravitational torque acting on the rod is: MY ANSWER - I said that it had 0 (zero) torque. Is this correct?
Now, the rod is given a nudge to the right, so that it starts to fall with nearly zero initial speed. Ignore air resistance.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is: MY ANSWER - I said that the torque was 1/2MgL. Is this correct or is the coefficient wrong?
3) What is the final angular velocity of the rod just before it hits the floor? Express the final answer in terms of variable L, M, g, and mathematical constants. (Hint: Use conservation of energy)
MY ATTEMPT - Conservation of energy is expressed by: (U_grav)i = (K_trans+K_rot)f ... where do I go from there?
1) When the rod is balanced vertically, the gravitational torque acting on the rod is indeed zero. Torque is defined as the product of force and the distance from the point of rotation. In this case, since the rod is perfectly vertical, the force due to gravity acts directly downwards along the rod's center of mass, passing through the hinge. The distance from the point of rotation (hinge) to the line of action of the force is zero, which means the torque is zero. So, your answer is correct.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is not 1/2MgL. The correct torque can be found by considering the position of the center of mass of the rod when it reaches that point. The rod will rotate about the hinge until the center of mass reaches the floor. At this moment, the torque due to gravity will be at its maximum.
To calculate this torque, we need to determine the perpendicular distance from the center of mass of the rod to the hinge. Since the rod is given to have a uniform density and thickness, its center of mass will be located at the geometric center of the rod, which is L/2 away from the hinge. So, the torque just before the rod hits the floor will be MgL/2.
Therefore, your answer is incorrect. The torque is not 1/2MgL, but rather MgL/2.
3) To determine the final angular velocity of the rod just before it hits the floor, we can use conservation of energy. The initial potential energy of the rod is MgL, and the final total energy is the sum of the translational kinetic energy and the rotational kinetic energy.
Using conservation of energy:
(MgL) = (1/2)(M)v^2 + (1/2)(I)ω^2
Where v is the linear velocity of the rod just before it hits the floor, I is the moment of inertia of the rod about the hinge (which can be calculated as 1/3 ML^2), and ω is the angular velocity.
From here, we need to express the linear velocity in terms of the angular velocity. Since the rod is rotating about the hinge, the linear velocity of any point on the rod can be related to its distance from the hinge by v = ωr, where r is the distance of the point from the hinge.
Considering the geometric arrangement, the far end of the rod (L/2 away from the hinge) will have the maximum linear velocity just before hitting the floor. So, v = ω(L/2).
Now, we can substitute this into the conservation of energy equation:
(MgL) = (1/2)(M)(ω(L/2))^2 + (1/2)(1/3)(ML^2)ω^2
Simplifying:
MgL = (1/8)Mω^2L^2 + (1/6)Mω^2L^2
Combining like terms:
MgL = (7/24)Mω^2L^2
Canceling the mass (M) from both sides:
g = (7/24)ω^2L
Simplifying further:
ω^2 = (24/7)g/L
Taking the square root of both sides:
ω = √(24g/7L)
So, the final angular velocity of the rod just before it hits the floor is √(24g/7L).