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=
(a) The torque at point B is given by: τB = (mass) * (acceleration due to gravity) * (distance from B to center of mass).
The mass of the stick is given as "m".
The distance from B to the center of mass is half the length of the stick, which is l/2.
Therefore, the torque at point B is: τB = m * g * (l/2).
Answer: τB = m * g * (l/2).
(b) The moment of inertia of a uniform stick rotating about one end is given by: I = (mass) * (length)^2 / 3.
The mass of the stick is given as "m".
The length of the stick is given as "l".
Therefore, the moment of inertia about point B is: I = m * (l^2) / 3.
The torque at point B is related to the moment of inertia by the equation: τB = I * α, where α is the angular acceleration.
Therefore, α = τB / I = (m * g * (l/2)) / (m * (l^2) / 3).
Simplifying, α = 3g / (2l).
Answer: α = 3g / (2l).
(c) The vertical acceleration of the center of mass is equal to the angular acceleration multiplied by the distance from the center of mass to point B.
The distance from the center of mass to point B is half the length of the stick, which is l/2.
Therefore, the vertical acceleration of the center of mass is: a = α * (l/2).
Substituting the value of α from part (b), a = (3g / (2l)) * (l/2).
Simplifying, a = 3g / 4.
Answer: a = 3g / 4.
(d) The vertical component of the hinge force at point B is equal to the mass of the stick multiplied by the vertical acceleration of the center of mass.
The mass of the stick is given as "m".
The vertical acceleration of the center of mass is given as 3g / 4 (from part (c)).
Therefore, the vertical component of the hinge force at point B is: N = m * (3g / 4).
Answer: N = (3mg) / 4.
To answer these questions, we need to apply the principles of rotational motion and the concept of equilibrium. Let's break down each part step by step:
(a) To find the magnitude of the torque (τB) about the end B at the edge of the table, we need to consider the forces acting on the stick at that point. In this case, the only force acting on the stick is the weight of the stick. The weight acts vertically downward through the center of mass of the stick, which is at the midpoint.
The torque about point B is calculated by multiplying the force (weight) by the lever arm, which is the distance from the point of rotation (B) to the center of mass of the stick (midpoint).
The weight of the stick can be calculated as the mass (m) multiplied by the acceleration due to gravity (g). The lever arm is half the length of the stick (l/2).
Therefore, the magnitude of the torque (τB) about point B can be calculated as:
τB = weight * lever arm = (m * g) * (l/2) = (mgl) / 2
So, the magnitude of the torque τB is (mgl) / 2.
(b) To find the magnitude of the angular acceleration (α) about the end B at the edge of the table, we need to consider the moment of inertia (I) of the stick about its end B and the torque (τB) acting on it.
The moment of inertia of a uniform stick about one end is (1/3) * m * l^2.
The torque (τB) acting on the stick is the net torque, which is equal to the moment of inertia (I) multiplied by the angular acceleration (α).
Thus, τB = I * α
Substituting the values, we have:
(mgl) / 2 = [(1/3) * m * l^2] * α
Now, we can solve for α:
α = [(mgl) / 2] / [(1/3) * m * l^2]
= (3gl) / (2 * l^2)
= 3g / (2l)
So, the magnitude of the angular acceleration α is 3g / (2l).
(c) To find the magnitude of the vertical acceleration (a) of the center of mass, we can use Newton's second law for rotational motion. This law states that the net torque acting on an object is equal to the moment of inertia times the angular acceleration.
In this case, the net torque acting on the stick is the torque (τB) about the end B, and the moment of inertia (I) of the stick about its center of mass is (1/12) * m * l^2.
Therefore, τB = I * α
Substituting the values, we have:
(mgl) / 2 = [(1/12) * m * l^2] * a
Now, we can solve for a:
a = [(mgl) / 2] / [(1/12) * m * l^2]
= (6g) / (l^2)
So, the magnitude of the vertical acceleration a of the center of mass is 6g / (l^2).
(d) The magnitude of the vertical component of the hinge force (N) at B can be found by considering the equilibrium of forces acting on the stick.
In this case, we have two forces acting on the stick: the weight acting vertically downward and the vertical component of the hinge force (N) acting vertically upward.
Since the stick is in equilibrium, the sum of the vertical forces must be zero.
Therefore, N - weight = 0
Substituting the value of the weight (m * g), we have:
N - m * g = 0
Hence, N = m * g
So, the magnitude of the vertical component of the hinge force (N) at B is m * g.
Prob # 7
ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/two-dimensional-rotational-motion/two-dimensional-rotational-dynamics/MIT8_01SC_problems21_soln.pdf