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. zero is correct
2. yes, any force up from the floor has no moment arm
3. change in Pe = m g h = (1/2) m g L
Ke = (1/2) m v^2 + (1/2) I w^2
I = (1/12) m L^2
v = (L/2)w
so
Ke = (1/2)m(L^2/4)w^2 + (1/2)(1/12)m L^2 w^2
= (m L^2 w^2) (1/8 + 1/24)
= m L^2 w^2 (1/6)
so
(1/2) m g L = m L^2 w^2 /6
w^2 = 3 g /L
w = sqrt (3 g/L)
1) Your answer that the gravitational torque acting on the rod is 0 is correct while the rod is balanced vertically. To understand why, let's consider the definition of torque.
Torque is the measure of the tendency of a force to rotate an object about an axis. It is given by the product of the force applied and the perpendicular distance from the axis of rotation to the line of action of the force.
In this case, when the rod is balanced vertically, the force of gravity acts directly downward through the center of mass of the rod. Since the gravitational force is acting along the axis of rotation (the hinge), the perpendicular distance from the axis to the line of action of the force is zero. Therefore, the torque due to gravity is zero.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is indeed not zero. As the rod starts to fall, its center of mass will move to a position lower than the hinge. This creates a torque due to gravity that causes the rotation of the rod.
To calculate this torque, we need to determine the perpendicular distance from the axis of rotation (the hinge) to the line of action of the gravitational force with the center of mass of the rod. This distance is half the length of the rod since the rod is balanced vertically. So, the torque due to gravity is given by:
Torque = Mass * gravitational acceleration * perpendicular distance from the axis of rotation
T = M * g * (L/2)
Therefore, your answer of 1/2MgL for the torque just before the rod hits the floor is correct.
3) To find the final angular velocity of the rod just before it hits the floor, you can use the conservation of energy principle.
The initial potential energy of the rod is purely gravitational and given by U_grav_i = M * g * L.
At the final moment just before the rod hits the floor, all its initial potential energy will be converted into kinetic energy. The total kinetic energy of the system consists of the translational kinetic energy of the center of mass and the rotational kinetic energy due to the spinning motion of the rod.
Now, the translational kinetic energy K_trans is given by 1/2 * M * v^2, where v is the speed of the center of mass just before hitting the floor. However, you mentioned that the rod starts to fall with nearly zero initial speed. Therefore, the translational kinetic energy can be neglected.
The rotational kinetic energy K_rot is given by 1/2 * I * w^2, where I is the moment of inertia and w is the angular velocity.
Since the rod is thin with uniform density, its moment of inertia can be approximated as 1/3 * M * L^2 (derived from the moment of inertia of a thin rod about one end).
Now applying conservation of energy:
U_grav_i = K_trans_f + K_rot_f
M * g * L = 0 + 1/2 * (1/3 * M * L^2) * w^2
Simplifying:
gL = (1/6) * L^2 * w^2
Squaring both sides and canceling out L:
g = (1/6) * w^2
Solving for w:
w^2 = 6g
w = sqrt(6g)
So, the final angular velocity of the rod just before it hits the floor is sqrt(6g), expressed in terms of variable L, M, g, and mathematical constants.