ew History Bookmarks Window Help Study 1 Guided Solutions and Study Help Chegg.c

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ew History Bookmarks Window Help Study 1 Guided Solutions and Study Help Chegg.com Ch.12:Rotation of a Rgisd Boty Part2 Problem 12.32 Problem 12.32 Part A A car tire is 61.0 cm in diameter. The car is traveling at a speed of 16.0 m/s What is the tire's rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units Pan My Answers Give Up Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units Value Units My Answers Give Up Part C What is the speed of a point at the bottom edge of the tre? Express your answer as an integer and include the appropriate units 1 pt 12. Problem 12.62 8 9

Explanation / Answer

First thing to do here is to convert the tyre diameter to 'metres' so 61cm = 0.61 metres

Then find the circumference of the tyre - Pi x Diameter i.e. 3.142 x 0.61 = 1.91662 metres

So if the car is travelling at 16 metres every second then this divided by the circumference of the tyre will give the frequency of rotation.

a) 16/1.91662 = 8.35 rps to 2 decimal places (note this is revs per 'second' and we need rpm)

So 8.35 x 60 = 501 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 16 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)

so,

B) top faster by a factor of two = 32 m/s
C) bottom is always zero because of the static frictional force = 0m/s

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