## Force to hold a cylinder about its axis with Center of Gravity offset from the center

This is a work related problem. I was calculating the amount of force required to hold a cylinder in a position when its Center of gravity is not at the center.

Problem Background
• Assume a 60' long cylinder.
• Circumferential it is made of 8 sections.
• For maintenance purpose, one section is completely removed (Side View shown in Picture3 & front view shown in Picture2). Hence the shift in center of gravity. (Shown in Picture2)
• This entire arrangement is rotated with the help of gear and pinion arrangement. (Picture1.)

To find :
Find Pt in Picture1 (Pt is the counter force)

Given Data:
• Bigger Circle : Gear
• Smaller Circle : pinion
• Center of Gravity CG point is 14.125 inches to the left from the vertical center line
• Radius of Gear = 144”
• Radius of pinion = 10.75”
• Reactions Rb and Rc are supports. Angle Rb-O-G = Angle Rc-O-G= 30 degrees
• Ra=1,111,887.5#

Attempt 1:
• Finding moments about point O.
• Therefore since direction of reactions Rb and Rc are in the line of action of force, they wouldn't have any moment about the point ‘O’ …. Correct me if I am wrong

Implies, Pt= (1111887.5*14.125)/144= 109065.3 #

My Doubts :
• Is this approach correct ? and is this the value of the force required to hold the cylinder
• Or do I have to consider the reactions Rb and Rc too ? If that is the case I shall end up having three unknowns, namely Rb,Rc and Pt.
Attached Thumbnails

 PhysOrg.com physics news on PhysOrg.com >> Study provides better understanding of water's freezing behavior at nanoscale>> Soft matter offers new ways to study how ordered materials arrange themselves>> Making quantum encryption practical
 Recognitions: Homework Help Science Advisor The supports are rollers, turning freely, yes? Then your calculation looks right.

 Quote by haruspex The supports are rollers, turning freely, yes? Then your calculation looks right.
Yes. Free Rolling !