A uniform stick of mass m and length l is suspended horizontally with end B at the edge of a table as shown in the diagram, and the other end A is originally held by hand.
The hand at A is suddenly released.
At the instant immediately after the release:
(a) What is the magnitude of the torque (τB) about the end B at the edge of the table?
Express your answer in terms of m, l and acceleration due to gravity g as needed (enter m for m, l for l and g for g).
τB=
(b) What is the magnitude of the angular acceleration α about the end B at the edge of the table?
Express your answer in terms of m, l and acceleration due to gravity g as needed (enter m for m, l for l and g for g).
α=
(c) What is the magnitude of the vertical acceleration a of the center of mass?
Express your answer in terms of m, l and acceleration due to gravity g as needed (enter m for m, l for l and g for g).
a=
(d) What is the magnitude of the vertical component of the hinge force (N) at B ?
Express your answer in terms of m, l and acceleration due to gravity g as needed (enter m for m, l for l and g for g).
N=
To answer these questions, we need to analyze the forces and torques acting on the stick after it is released. Let's break down each part of the question.
(a) The magnitude of the torque about point B can be found using the formula for torque: τ = r × F, where τ is the torque, r is the position vector from the pivot point to the line of action of the force, and F is the force.
In this case, the force acting on the stick is the weight of the stick, which is mg, and the position vector r is the perpendicular distance from point B to the line of action of the force. Since the stick is suspended horizontally, this distance is l/2. Therefore, the torque about point B is given by:
τB = r × F = (l/2) × (mg) = mgl/2
So, the magnitude of the torque about point B is mgl/2.
(b) The angular acceleration about point B can be found using the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
For a thin rod rotating about one end, the moment of inertia is given by I = (1/3)ml^2. Substituting the value of torque (mgl/2) into the equation, we can solve for α:
mgl/2 = (1/3)ml^2 α
Simplifying the equation, we find:
α = (3g)/2l
So, the magnitude of the angular acceleration about point B is (3g)/2l.
(c) The vertical acceleration of the center of mass can be found using the equation F = ma, where F is the net force acting on the rod, m is the mass of the rod, and a is the acceleration.
In this case, the only force acting on the rod is the vertical component of the hinge force, N. Therefore, the net force is given by:
F = N - mg
Setting this net force equal to ma, we have:
N - mg = ma
Simplifying the equation, we find:
a = (N - mg)/m
So, the magnitude of the vertical acceleration of the center of mass is (N - mg)/m.
(d) To find the magnitude of the vertical component of the hinge force at B, we can use the equation of torque equilibrium:
∑τ = 0
Since the stick is in rotational equilibrium, the sum of the torques about any point must be zero. Taking point B as the pivot, the only torque acting on the stick is the torque due to the weight of the stick (mg) about point B, which is equal to mgl/2. Therefore, the hinge force at B must provide an equal and opposite torque to maintain equilibrium.
So we have:
τB = N × 0 = mgl/2
Simplifying the equation, we find:
N = (mgl/2)/0
Notice that the denominator is zero, which means that the hinge force at B will be undefined. This indicates that there is no vertical component of the hinge force at point B.
Therefore, the magnitude of the vertical component of the hinge force at B is undefined (N = undefined).