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) While the rod is balanced vertically, the gravitational torque acting on it is indeed zero. This is because the force of gravity acts vertically downwards through the center of mass of the rod, and there is no lever arm to create a torque.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is not 1/2MgL. The correct value can be determined by considering the point of rotation (the hinge). Let's assume the hinge is at the bottom of the rod.
The gravitational torque acting on the rod is given by the equation: τ = r x F, where τ is the torque, r is the lever arm, and F is the force. In this case, the force is the weight of the rod, which is equal to Mg. The lever arm is the distance from the hinge to the center of mass of the rod, which is L/2.
So, the gravitational torque is τ = (L/2) x Mg = (LMg)/2 = 1/2LMg.
Therefore, the correct answer is 1/2LMg.
3) To find the final angular velocity of the rod just before it hits the floor, we can use the conservation of energy.
The initial potential energy of the rod is U_grav = MgL/2 (considering the center of mass at a height L/2).
The final kinetic energy of the rod can be divided into translational kinetic energy (K_trans) and rotational kinetic energy (K_rot). The translational kinetic energy is given by K_trans = (1/2)Mv^2, where v is the linear velocity of the rod just before it hits the floor. The rotational kinetic energy is given by K_rot = (1/2)Iω^2, where I is the moment of inertia of the rod about the axis of rotation (taken as the hinge) and ω is the angular velocity.
Since the rod starts at rest and falls with nearly zero initial speed, we can assume its initial linear velocity is negligible. Therefore, K_trans ≈ 0.
The moment of inertia of a thin rod about an axis perpendicular to its length and passing through its center of mass is given by I = (1/12)ML^2.
Using the conservation of energy equation (U_grav)i = (K_trans+K_rot)f, we can substitute the values:
MgL/2 = (0 + (1/2)(1/12)ML^2ω^2).
Simplifying the equation, we get:
g = (1/24)Lω^2.
Solving for ω, we have:
ω^2 = (24g)/L.
Therefore, the final angular velocity of the rod just before it hits the floor is ω = √((24g)/L).
1) While the rod is balanced vertically, the gravitational torque acting on the rod is indeed zero. This is because the gravitational force acts directly along the rod's axis, causing the torque arm (perpendicular distance from the axis of rotation to the line of action of the force) to be zero. Thus, the torque is given by the equation torque = force x torque arm. Since the torque arm is zero, the torque is zero as well.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is also zero. As the rod falls, its center of mass moves downward, causing the gravitational force to apply a torque in the opposite direction due to the fixed hinge point. However, in order to maintain rotational equilibrium, there must be an equal and opposite torque present. This additional torque is provided by the normal force acting at the hinge, which counteracts the gravitational torque. Therefore, the net gravitational torque is zero.
Your answer of 1/2MgL for the gravitational torque just before the rod hits the floor is incorrect. The correct answer is zero, as explained above.
3) To find the final angular velocity of the rod just before it hits the floor, we can use the conservation of energy principle. The initial potential energy (U_grav)i is equal to the sum of the final translational kinetic energy (K_trans)f and final rotational kinetic energy (K_rot)f.
The initial potential energy, given by U_grav = Mgh, where h is the height from which the rod falls (equal to the length L of the rod in this case).
The final translational kinetic energy, given by K_trans = (1/2)Mv^2, where v is the final velocity of the center of mass.
The final rotational kinetic energy, given by K_rot = (1/2)Iω^2, where I is the moment of inertia of the rod about its center of mass and ω is the final angular velocity.
Since the rod starts from rest, it has no initial translational or rotational kinetic energy (K_trans+i = 0, K_rot+i = 0).
Therefore, we have U_grav = K_trans+f + K_rot+f
Substituting the relevant equations:
Mgh = (1/2)Mv^2 + (1/2)Iω^2
The moment of inertia of a thin rod rotating about its center of mass is given by I = (1/3)ML^2.
Substituting this into the equation, we have:
Mgh = (1/2)Mv^2 + (1/2)(1/3)ML^2ω^2
Rearranging the equation, we get:
Mgh - (1/2)Mv^2 = (1/6)ML^2ω^2
The final velocity v of the center of mass just before the rod hits the floor is approximately zero (as given in the question). Therefore, we can neglect the (1/2)Mv^2 term.
Now we can solve for the final angular velocity ω:
Mgh = (1/6)ML^2ω^2
Simplifying the equation:
ω^2 = (6gh) / L^2
Taking the square root of both sides:
ω = √(6gh) / L
So, the final angular velocity of the rod just before it hits the floor is √(6gh) / L, where g is the acceleration due to gravity.