Sir,
My question is as below:-
Three particles a b & c of mass m are connected to each other by three massless rigid rods to from a rigid, equilateral triangular body of side 'l'.This body is is placed on a horizontal frictionless table (x-y plane) an is hinged to it at the point A so that it can move without friction about the vertical axis through A as shown in figure. The body is set into rotational motion on the table about A with a constant angular velocity w[omega)
a)Find the magnitude of the horizontal force exerted by hinge on the body.
b) At the time T when the side BC is parallel to the X-axis, force F is applied on B along BC as shown. Obtain X-component and Y-component of the force exerted by the hinge on the body immediately after time T.
To find the magnitude of the horizontal force exerted by the hinge on the body, we can start by analyzing the forces acting on the body.
a) Magnitude of Horizontal Force:
Since the body is rotating about point A with a constant angular velocity (w), there must be a net external torque acting on the system. However, since the system is on a frictionless table, there is no external torque acting on it. Therefore, the angular momentum of the system about point A must be conserved.
The angular momentum (L) of a particle rotating about an axis is given by the equation:
L = I * w
Where:
L = Angular momentum
I = Moment of inertia
w = Angular velocity
In this case, since the body is an equilateral triangle with mass (m), the moment of inertia can be calculated as:
I = (1/6) * m * l^2
Where:
I = Moment of inertia
m = Mass of each particle
l = Length of each side of the equilateral triangle
Since the body is rotating, we can calculate the angular momentum of each particle separately and then sum them up for the entire system.
The angular momentum of each particle can be calculated as:
L_particle = m * r * v
Where:
L_particle = Angular momentum of the particle
m = Mass of the particle
r = Distance between the particle and axis of rotation (point A)
v = Linear velocity of the particle
In an equilateral triangle, the distance between each particle and the center of mass is given by:
r = l / sqrt(3)
For a rotating body in circular motion, the linear velocity of each particle can be calculated using the equation:
v = r * w
Substituting the values and calculating for each particle, we get:
L_particle = m * (l / sqrt(3)) * (l / sqrt(3)) * w = m * l^2 * w / 3
Since the body consists of 3 particles, the total angular momentum of the system is:
L_system = 3 * L_particle = 3 * m * l^2 * w / 3 = m * l^2 * w
Now, since angular momentum is conserved, the initial angular momentum of the system (before applying the force) must be equal to the final angular momentum of the system (after applying the force).
At the initial time when the body is set into rotational motion, the side of the triangle is perpendicular to the X-axis. This means that the particles have initial velocities in the Y-direction:
v_initial = l * w
Therefore, the initial angular momentum of the system is:
L_initial = 3 * m * (l / sqrt(3)) * (l * w) = m * l^2 * w
Since angular momentum is conserved, the final angular momentum of the system (after the side BC becomes parallel to the X-axis) is also:
L_final = m * l^2 * w
Thus, we can conclude that the magnitude of the horizontal force exerted by the hinge on the body is zero. This is because there is no external torque acting on the system, and the angular momentum of the system is conserved.
b) X-Component and Y-Component of Force:
To find the X-component and Y-component of the force exerted by the hinge on the body immediately after time T, we need to consider the force F applied on particle B along BC.
When the side BC becomes parallel to the X-axis, the angle between the force F and the X-axis is 60 degrees. Therefore, the X-component of the force exerted by the hinge can be calculated by resolving the force F on the X-axis:
Force_X = F * cos(60)
Similarly, the Y-component of the force exerted by the hinge can be calculated by resolving the force F on the Y-axis:
Force_Y = F * sin(60)
Substituting the value of cos(60) = 1/2 and sin(60) = sqrt(3)/2, we get:
Force_X = (1/2) * F
Force_Y = (sqrt(3)/2) * F
Therefore, the X-component of the force exerted by the hinge on the body immediately after time T is (1/2) times the magnitude of the applied force F, and the Y-component of the force exerted by the hinge on the body immediately after time T is (sqrt(3)/2) times the magnitude of the applied force F.