a thin square homogeneous plate, AB=BC=CD=DA=2 m density 78 kn/m2 and thickness 10mm is suspended. A and B is fixed. if support a breaks suddenly calculate (1)instaneneous accelration of D,(2) instataneous angular acceleration op plate and (3) instateneous reaction at B.
To calculate the instantaneous acceleration of point D, we can use the equation for linear acceleration:
Acceleration (a) = (Change in velocity) / (Time taken)
However, since we do not have the velocity of point D, we need to find it first. We can use the principle of conservation of angular momentum to calculate the angular velocity of the plate when support A breaks.
The equation for conservation of angular momentum is:
Initial Angular Momentum = Final Angular Momentum
The initial angular momentum is 0, as the plate is initially at rest. The final angular momentum can be calculated as the product of moment of inertia (I) of the plate and the angular velocity (ω).
Now, let's move step by step to find the answers:
1. Instantaneous acceleration of point D:
To calculate the instantaneous acceleration at point D, we need to determine the initial angular velocity of the plate. For this, we can use the conservation of angular momentum:
Initial Angular Momentum = Final Angular Momentum
Since the initial angular momentum is 0, we can set it equal to the final angular momentum:
0 = I * ω
To find the moment of inertia (I) of the square plate, we can use the formula for the moment of inertia of a thin square plate about an axis passing through its center:
I = (1/12) * M * (AB^2)
Where M is the mass of the plate, and AB is the side length of the square plate.
Given:
Density (ρ) = 78 kN/m^2
Thickness (t) = 10 mm = 0.01 m
Side length (AB) = 2 m
First, let's calculate the mass of the plate:
Mass (M) = Density * Area * Thickness
Area = AB * AB
Area = 2 m * 2 m = 4 m^2
Mass (M) = 78 kN/m^2 * 4 m^2 * 0.01 m
Mass (M) = 3.12 kN
Now, let's calculate the moment of inertia:
I = (1/12) * M * (AB^2)
I = (1/12) * 3.12 kN * (2 m)^2
I = 0.26 kN * m^2
Since the mass is already given in kilonewtons (kN), the moment of inertia is also in kilonewton-meters squared (kN * m^2).
Now we can solve for the final angular velocity (ω) using the equation:
0 = I * ω
ω = 0
Since the final angular velocity is 0, the instantaneous acceleration at point D will also be 0.
Therefore, the instantaneous acceleration at point D is 0.
2. Instantaneous angular acceleration of the plate:
Since the instantaneous acceleration at point D is 0, we can conclude that the angular acceleration of the plate is also 0. This is because the plate is rotating about an axis of rotation passing through the center of mass, and the point at D is at the same distance from the axis of rotation as the center of mass.
Therefore, the instantaneous angular acceleration of the plate is 0.
3. Instantaneous reaction at point B:
To calculate the instantaneous reaction at point B, we can use Newton's second law of rotation:
Torque (τ) = Moment of Inertia (I) * Angular Acceleration (α)
The torque acting on the plate is due to the weight of the plate acting at its center of mass. The weight (W) can be calculated as the product of mass (M) and gravitational acceleration (g):
Weight (W) = Mass (M) * Gravitational Acceleration (g)
Given:
Mass (M) = 3.12 kN
Gravitational Acceleration (g) = 9.8 m/s^2
Weight (W) = 3.12 kN * 9.8 m/s^2
Weight (W) = 30.576 kN
Now, let's calculate the torque acting on the plate:
Torque (τ) = Weight (W) * Distance (d)
The distance (d) from point B to the center of mass of the plate is half the side length of the square plate (AB/2).
Distance (d) = AB/2 = 2 m / 2 = 1 m
Torque (τ) = 30.576 kN * 1 m
Torque (τ) = 30.576 kNm
Now, we can solve for the instantaneous reaction at point B by equating the torque and the product of moment of inertia and angular acceleration:
Reaction at B = Torque (τ) / Moment of Inertia (I)
Reaction at B = 30.576 kNm / 0.26 kNm^2
Reaction at B = 117.6 kN
Therefore, the instantaneous reaction at point B is 117.6 kN.