SOLUTION: Two friction wheels are of diameter 20 mm and 200 mm respectively. They touch at P and rotate without slipping.
Calculate the number of turns made by the small wheel when the lar
Question 1206775: Two friction wheels are of diameter 20 mm and 200 mm respectively. They touch at P and rotate without slipping.
Calculate the number of turns made by the small wheel when the large wheel rotates through 60°. Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52767) (Show Source): You can put this solution on YOUR website! .
The invariant is the product , the radius times the angle.
So, we write
= ,
or
= 100*60.
From this equality, we find
= = 10*60 = 600 degrees = 360 degrees + 240 degrees,
so the small wheel will make one full rotation and 2/3 of the other rotation.
The ratio of diameters of the two wheels is 20:200, or 1:10.
The wheels touch on their circumferences, which are also in the ratio 1:10.
Since the wheels rotate without slipping, the ratio of the angles through which they rotate are in the ratio 1:10, with the smaller wheel of course rotating faster.
The large wheel rotates through an angle of 60 degrees; the small wheel rotates through an angle of 10*60 degrees = 600 degrees.
The number of rotations the small wheel makes (360 degrees per revolution) is 600/360 = 5/3.