﻿ Step 13.3: A Rotating Rocker

# Step 13.3: A Rocker

### A Rocker – a Crank

We connect a Linear-Motion FB [a Clock FB] to rotate the Rocker with a constant angular velocity.

This Step helps to understand:

 1 'Centripetal Acceleration and Force' that is a result of the circular motion of the Centre-of-Mass
 2 'Superposition of Gravitational and Centripetal Force'
 3 Why the 'Moment of the Centripetal Force' equals zero.

 STEP 1: Delete the Spring Force FB or Edit the Spring-Force FB and make the Spring-Rate = 0 N/mm
 STEP 2: Add a Linear-Motion FB
 STEP 3: Connect a wire from the output-connector of the Linear-Motion FB to the input-connector of the Motion-Dimension FB.

The Machine Speed Setting is 60RPM, or 1Cycle/second. 1Cycle/second = 2π radians/second.

The mechanism below shows the forces that act at the Pin-Joint. Addition of Vertical Forces acting on the Rocker [Point 2] : [+ve].

∑FV=0 : R2V[N] - 1[kg]*9.807[m/s/s] = 0;  R2V = 9.807N [upwards]

Addition of Horizontal Forces acting on the Rocker [Point 2] : [ +ve].

∑FV=0 : +R2H[N] + 1[kg]*0.1[m]*[2π]2[1/s/s] = 0;  R2H = -3.948N [to the left]

Moments are:

 • An Anti-clockwise Moment that acts on the Rocker is 0.98Nm - as before.
 • The Centripetal Force acts through the centre of rotation. It does not apply a moment to the Arm. Hence, the Moment is the same. The image to the left is the same as the image above.

I have added to the image the Horizontal and Vertical Force Vectors that ACT-ON the Rocker

The two components are:

 1 Reaction to the Gravitational Force - it acts on the Rocker. Equal to 9.81N
 2 Centripetal Force acts on the Rocker to the left. Equal to 3.948N

The forces are perpendicular[⊥] when the Rocker is horizontal. Hence, we can use Pythagoras, to give 10.571N

Ftotal = √(Fg2 + Fc2) (Total Force = SQRT(SQR(Gravitation Force) + SQR(Centripetal Force)) = SQRT((m.g)2 + (m.r.ω2)2)

Total Force = √((1*g)2 + (1*0.1*(2π)2)2) = √(9.812 + 3.9482) = 10.571N

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