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# Uniform Circular Motion Help?

1.A car tire is spinning at a constant rate of 660rpm when the cars linear speed is 22m/s along the road.Determine (a)the frequency of rotation of the tire,(b)the period of rotation,and(c)the radius of the tire. NOTE pi=3.14

2. The earth goes around the sun once per year(as we know).Calculate the linear speed of the earth as it goes around the sun.The average distance between the two is 150 million kilometers.Suppose the orbit is a perfect circle although it is not

with solution please thanks

### 1 Answer

- Pramod KumarLv 71 decade agoFavorite Answer
Q1.

Let the radius of the tire = R meter

Hence circumference of the tire = 2 π R m

It means the tire covers a distance of ( 2 π R ) m in every revolution ( spin is a wrong term used here ).

The question further says " The tire revolves at a constant rate of 660rpm "

Hence the distance covered in meter per min = 660 ( 2 π R )

=> Distance covered per second ( ie linear speed of the tire ) = 660 ( 2 π R ) / 60 = 22 π R m/s

But according to the problem this speed = 22 m/s

=> 22 π R = 22

=> R = ( 1 / π ) m = ( 1 / 3.14 ) m = 0.31847 m = 31.847 cm ............. Answer to part c

If PERIOD OF ROTATION ie time taken for one rotation = T, then

=> T = 1 / 660 min = (60/660) = 1/11 s .......... Answer to Part b

Frequency of Rotation f = 1 ? T = 11 rev/s ................. Answer to Part a

............................................................................

Soln Prob : 2

Radius of Rotation = R = 150 million kilometers

Hence the linear speed of the earth ( V ) = 2 π ω R where ω is the angular speed of the earth.

V = 2 π ( 1 ) ( 150 ) million kilometer per year.

=> V = ( 2 π x 150 x 1000000 ) / ( 365x24 ) = ( 0.107588789 x 1000000 ) km/hr

=> V = 107588.789 km / hr = 29.888577486 km/s = 29888.577486 m/s

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