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# A car tire is 54 cm in diameter. The car is traveling at a speed of 10 m/s.?

A car tire is 54 cm in diameter. The car is traveling at a speed of 10 m/s.

(a) What is the tire's rotation frequency, in rpm?

(b) What is the speed of a point at the top edge of the tire?

(c) What is the speed of a point at the bottom edge of the tire?

### 3 Answers

- 10 years agoFavorite Answer
First thing to do here is to convert the tyre diameter to 'metres' so 54cm = 0.54 metres

Then find the circumference of the tyre - Pi x Diameter i.e. 3.142 x 0.54 = 1.69668 metres

So if the car is travelling at 10 metres every second then this divided by the circumference of the tyre will give the frequency of rotation.

a) 10/1.69338 = 5.89 rps to 2 decimal places (note this is revs per 'second' and we need rpm)

So 5.89 x 60 = 353.63 rpm to 2 decimal places

Now for b) & c) Since we do not mention angular 'velocity' it is assumed that the points indicated are both moving at 10 m/s and in the same general direction as the car, ignoring frictional effects and slippage. Indeed it is the Tyre's that are transmitting the force that develops the forward speed in this case. However, if we consider the 'single point' in contact with the road, then at that very instant the point is stationary due to the grip or friction. This point quickly moves due to the induced rotational forces and is no longer the point at the bottom of the tyre. Conversely for the top point.

A different analogy to consider is a 'tank track' here, the top part of the track moves at twice the speed of the tank whereas the bottom of the track remains stationary (i.e. remains in contact with the road or whatever it's resting on) - strange but true.

Hope this is of some help.

- 6 years ago
B) top faster by a factor of two = 20m/s

C) bottom is always zero because of the static frictional force = 0m/s