# Rotation of Planetary gears about a fixed Sun gear

This coursework question I am faced with is very specific and quite difficult to explain so I have attached the question. The writing on the side is not that important so dont try and make sense of it.

The question asks to find the angular velocity of 2 planetary gears (which are connected by a quarter circle plate) rotating about a larger Sun gear. The plate is attached to the centre of the planetary gears at the corners where the arc and the radii meet, with the 2 radii converging at the centre of the Sun gear.

the mass of the 2 planetary gears are 2kg

the mass of the quarter circle is 6kg

the radius of the Sun gear is 0.15m

the radius of the planetary gears are 0.075m

things I worked out from that:

Radius of quarter circle = 0.225m

Centre of mass of the quarter circle = 0.135m from the pinnacle.

Initially, the quarter circle occupies quadrant 1 entirely, with the planetary gears connected at the corners.

The question asks to find the angular velocity of the planetary gears when the quarter circle moves through Pi/2 rad clockwise to occupy quadrant 4 entirely.

This is an energy conservation question, however I'm struggling with the relevant circular motions.

KE + PE + RE=constant

I found the centre of mass of the quarter circle and thus found the initial energy of the system but wasnt sure what to do next.

I want to find the final KE but I am not sure how to approach this. I know gravity is working tangentially to the Sun gear, but how do you incorporate that into the relevant KE equation?

#### Attachments

• 22.9 KB Views: 404

Related Engineering and Comp Sci Homework Help News on Phys.org
tiny-tim
Homework Helper
Welcome to PF!

Hi SugreF! Welcome to PF!
… I want to find the final KE but I am not sure how to approach this. I know gravity is working tangentially to the Sun gear, but how do you incorporate that into the relevant KE equation?
You don't …

KE is geometry, and has nothing to do with forces …

it's just 1/2 mvc.o.m.2 + 1/2 Ic.o.m.ω2, = 1/2 Ic.o.r.ω2