Question

The crank OA rotates in the vertical plane with a constant clockwise angular velocity omega knot of 4.5 rad/s. For the position where OA is horizontal calculate the force under the light roller B of the 10-kg slender bar AB.

Answer 1

Concepts and reason

Angular velocity:

The rate at which the angular displacement of a body varies with respect to time is angular velocity. It is denoted by and its unit is .

Acceleration:

The change in velocity with respect to time of an object is called acceleration. Its S.I unit is .

Newton’s second law:

It states that “the force applied to a body in a direction is directly proportional to the mass and acceleration of the body along that direction”.

Force of a moment:

The magnitude of the moment can be determined by the product of force and the perpendicular distance to the force and the reference point.

Instantaneous center:

It is a point in the plane of motion of an object at which the velocity of the object is negligible or zero. The object will appear to rotate around the instantaneous center.

Consider the Figure (1) shown below:

Here, the velocity of point A is , the velocity of point B is , velocity of instantaneous center is , position of A with respect to instantaneous center is , position of B with respect to instantaneous center is and the angular velocity of body is .

For the Figure (1), the instantaneous center is determined as the intersection of perpendicular lines drawn to the velocities of point A and point B.

Calculate the angular velocity of bar using instantaneous center method. From the definitions of accelerations, obtain the angular acceleration of bar AB. Then apply the equilibrium moment condition under Newton’s second law to determine the force under the light roller at B.

Fundamentals

Weight:

The weight of an object is the mass times the acceleration due to gravity on an object.

Write the expression for weight of an object.

Here, the mass of object is and the acceleration to gravity is .

Write the relation for Angular velocity .

Here, the constant velocity is , and the radius of the circular path is r.

The normal acceleration acting on an object in rotation is given as follows:

The tangential acceleration acting on an object is angular acceleration times the radius of curvature.

Here, the angular acceleration is and the radius of curvature for circular path is r.

Write the formula for mass moment of inertia of a slender rod.

Here, mass of slender rod is and the length of rod is .

Write the formula for the moment due to force.

Here, the force is and the perpendicular distance is .

Write the Newton’s second law equation.

Here, force is F, and the acceleration of mass is a.

Write the equilibrium moment under Newton’s second law about any reference point.

The summation of moments about a point is and summation of moments due to mass is .

General sign conventions of moment:

The moment is considered as positive in counter clockwise direction and moment is considered as negative in clockwise direction.

Calculate the velocity of end A for the crank OA.

The length of crank is and the angular velocity of crank is .

Substitute for and for .

Find the instantaneous center for slender bar AB as in Figure (2).

Here, the velocity of end B is .

Calculate the length x from Pythagoras formula.

Calculate the angular velocity of slender rod AB from Figure (2).

Substitute for and for .

Calculate the acceleration of point A.

Substitute for and for .

Calculate the normal acceleration component of point B with respect to A.

Substitute for and for .

Calculate the tangential acceleration component of point B with respect to A.

Substitute for .

Calculate the angular acceleration of bar AB.

Substitute for and for .

Consider the free body diagram for the slender bar AB as in Figure (3).

Here, force under roller is , weight of bar is , force at A in x direction is and force at A in y direction is .

Calculate the tangential acceleration by the mass at G with respect to A.

Substitute for , for and for .

Write the moment equilibrium condition about end A from Newton’s second law.

The summation of moments about A is and summation of moments due to mass is .

Substitute for .

Here, mass of slender bar is and acceleration due to gravity is .

Substitute for , for , for , for , for and for .

Ans:

The force under the light roller at B is .