Physics Quiz 10

[dellagd]

Originally Posted by Jared

I'm not sure how friction factors into this. A really wide wheel will have much more turning scrub than a thin or donut shaped wheel.

Here's what I've got so far:
I broke the wheel velocity into two components, one in the direction from the wheel to the center of the bot, and the other perpendicular to this one. The inward pointing one is all lost to friction. The other one has a magnitude of 0.707 * 2 = 1.414 feet/second.

The distance from the center to the wheel is 2.828 feet. This means one rotation is 2*pi*r=17.768 feet. This means it's rotating at 1.414/17.768 = 0.079 rotations per second.

I think I'm missing something eight the friction.

I'm not getting the friction part yet either, though I think you may have miscalculated the radius you used, 2.828 is the distance from wheel to wheel diagonally, making the rotation .159 rev/s. With the friction I'm guessing it will be less than this.

[seg9585]

I would think the coefficient of kinetic friction (and robot mass, for that matter) comes into play when calculating torque required to continue to rotate at the determined steady-state speed. But this was not asked.

[Bryce2471]

My first thought was 2rad/sec, but I'm pretty sure that's with 4 omni wheels. In order to solve this you need to know how much linear distance is traveled, when 2ft of tire tread is pulled by.

[Jared]

Oops. I change my answer to .158 rotations per second.

[FrankJ]

Just where do you define the tangential speed? Tangent to the circle subscribed through the wheels center line? if so rotation is 2ft/s* C/8.884ft=.2251rev/sec.

If "Tangential speed" is the slip between the wheel & the ground normal to the wheel axis, then life becomes more complicated & you need weight & friction. For instance a coefficient of 0 rotational speed is uncoupled from wheel speed.