A uniform square plate with side a=300 mm is hinged at points A and B to a clevis that rotates w

A uniform square plate with side a=300 mm is hinged at...

A uniform square plate with side $a=300 \mathrm{mm}$ is hinged at points $A$ and $B$ to a clevis that rotates with a constant angular velocity $\omega$ about a vertical axis. Determine $(a)$ the value of $\omega$ for which the plate forms a constant angle $\beta=60^{\circ}$ with the horizontal $x$ axis, $(b)$ the largest value of $\omega$ for which the plate remains vertical $\left(\beta=90^{\circ}\right) .$

Key Concepts

Rigid Body Equilibrium

For any rigid body, static equilibrium is achieved when both the net force and the net moment (torque) about any point are zero. This condition underlies the analysis of how gravitational forces and other external forces (or their equivalents in a rotating frame) act on a structured body, ensuring that it is neither translating nor rotating unexpectedly. This concept is crucial when assessing the balance of forces in systems with fixed supports or hinges.

Rotating Reference Frames and Fictitious Forces

When a system rotates with a constant angular velocity, analyzing its dynamics often involves adopting a rotating frame of reference where fictitious forces, such as the centrifugal force, must be introduced. These forces are not real in an inertial sense but are necessary to account for the observed equilibrium conditions in the rotating system. Understanding these fictitious forces is key to determining how the effective ‘gravity’ field is modified in a rotating environment.

Moment Balance and Lever Arm Concepts

Moments, or torques, result from forces acting at a distance from a pivot point and are the product of the force magnitude and its lever arm. For a body under rotational motion or with fixed pivot constraints, the balance of these moments determines its orientation. Analyzing how the gravitational force and inertial (centrifugal) force produce moments about the hinges demonstrates the interplay between a body's geometry and dynamic effects, which is vital for predicting equilibrium configurations.

Geometric Relationships in Rotational Equilibrium

The geometric arrangement of a rigid body—specifically, the relative positions of the center of mass, the forces acting on it, and the pivot or hinge points—directly influences the conditions for equilibrium. The orientation angle of the body is determined by how the effective gravitational field (a combination of real gravity and inertial effects) interacts with these geometrical parameters. This relationship is essential for understanding how variations in angular velocity can alter the body's stable configurations.

Transcript

00:01 Everyone here it is given a uniform semicircular plate of radius 120mm that is point 1 2 meter hinge at points a and b so this is the point a and this is the point b is clavis which rotate with constant angular velocity omega above the vertical axis determine omega calculate angular velocity omega for which platform form an angle of 50 degree with the horizontal axis.

00:54 Take the axis, see x -des -y -dice -z -z -z.

01:07 So this we will take x -des -dis and this will be the y -d -dice.

01:16 This angle is beta, so this would be also beta.

01:20 So we can resolve omega into two components.

01:23 You may write omega -2b.

01:26 Minus omega sine of beta i -d s plus omega cos of beta j -d -s now moment of inertia of the semicircular plate moment of inertia of semi -circular plate i x -dose is mr square by 4 and i y -dice mr square by 2 because of symmetry we can write symmetrically we can write by symmetry we can write i x dash by dash i y -daz z i z x -d x is 0 so angular momentum about c hc can be written as i x -des omega x -dose i -dice i -dice i -y -dose omega -y -dice substitute the value half mr square omega sine of veta this is negative i -dice half sorry this is 1 by 4 mr square omega cos of beta j s so this we can write 1 by 4 m r square omega minus sine of veta i dash plus 2 cos of beta j s now angular velocity of reference frame see x x x x x x x x x is is omega which is minus omega sine of beta i dash plus cos of omega cos beta j z so rate of change of angular momentum of the half plate about c will be rate of change of angular momentum with respect to reference frame c x x x plus omega cross at c.

05:35 This is zero because it is rotating with constant angular velocity.

05:41 Omega is minus omega sine of veta i.

05:49 Plus omega cos of beta j -d s cross product...

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