# Circular Motion of motorcyclist

1. Apr 27, 2007

### danago

A 50kg motorcyclist on a 100kg bike enters a tight curve of radius 30m at a speed of 20ms-1. The maximum frictional force the surface can provide is 2000N before the bike will skid uncontrollably.

• Will she make the curve safely?
• What will happen if she speeds up while on the curve?
• What would happen if she tried to slow down while on the curve?
• At what angle to the vertical does she lean her bike while rounding the corner?
• Why would this curve be safer if it was banked?

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a)
$$F_c=\frac{mv^2}{r}=\frac{150(20^2)}{30}=2000N$$

The centripetal force required to move in that circular path is 2000N, which is supplied by the friction, so she will JUST be able to travel that path.

b) If she speeds up, the centripetal force required to continue on that path exceeds 2000N, which cannot be supplied by the friction, so she will therefore skid out of control.

c) If she slows to below 20m/s, then the centripetal force required will be less than 2000N. The maximum friction from the ground is 2000, so she should still be able to maintain her circular path. Apparently, according to the answers, she will skid if she goes lower than 20m/s. Why is this? The answer says "Slowing down requires extra friction; she will skid"

Parts a and b im ok with, but im a little confused with part c. Any help would be great.

Thanks,
Dan.

2. Apr 27, 2007

### Sojourner01

Consider what has to happen to reduce the bike's speed. Kinetic energy must be lost through W=Fd between the tyres and the road. At the point where the bike is travelling 20ms-1 and requires additional friction to slow down, this will exceed the 2000N limit and the skid will occur.

3. Apr 27, 2007

### daniel_i_l

You could also look at the acceleration needed to change speeds. Since F=ma then a force is needed.