Month: January 2015

096 Circular Motion: Circling Nuts

So, is the required centripetal force larger for large radius or small radius?

Careful, you better not use F=mv2/r, since the tangential speed is different for nuts at different radius for this scenario. So it is not clear whether smaller r (and smaller v) requires larger (or smaller) F.

Use F=mrω2  instead, since in this scenario, all the nuts have the same angular velocity ω. Then it becomes clear that at the same ω, the outer nuts doing circular motion with a larger r require a larger centripetal force compared to the inner nuts. Outer nuts skid first.

In fact, if one is situated right at the middle, no centripetal force is required at all. This explains why the bean at the middle had the last laugh.

095 Atmospheric Pressure Mat

At 101 kPa, there is (0.30 x 0.30 x 101,000) = 9090 N, or about 900 kg worth of weight pressing down on the 30 cm by 30 cm pad. Because there is about 900 kg of air sitting on top of the pad!

But of course, we can’t expect the pad to lift 900 kg. Two reasons: (1) It is NOT a total vacuum under the pad. So the net downward force would be much less. (2) The pad itself (the rubber, handle knob, etc) would have been pulled apart long before that.