A circular coin of radius 1 cm is allowed to roll freely on the periphery over a circular disc of radius 10 cm. If the disc has no movement and the coin completes one revolution rolling on the periphery over the disc and without slipping, then what is the number of times the coin rotated about its centre ?
[amp_mcq option1=”10″ option2=”10.5″ option3=”11″ option4=”12″ correct=”option3″]
This question was previously asked in
UPSC CAPF – 2016
– The coin rolls on the periphery of the disc without slipping.
– The path traced by the center of the coin is a circle with radius R + r = 10 + 1 = 11 cm.
– When a smaller circle (radius r) rolls on the outside of a larger circle (radius R), the number of rotations the smaller circle makes about its own center for one complete revolution around the larger circle is given by the formula (R/r) + 1.
– In this case, R=10 cm and r=1 cm.
– Number of rotations = (10 cm / 1 cm) + 1 = 10 + 1 = 11.
– This formula accounts for the rotation due to the rolling action (distance rolled / circumference of coin) and the additional rotation due to the coin revolving around the large disc (one full revolution).
– The distance rolled by the coin along the edge of the disc is the circumference of the disc, which is 2 * pi * R = 2 * pi * 10 = 20 pi cm. The rotation due to this rolling is (20 pi) / (2 * pi * r) = 20 pi / (2 * pi * 1) = 10 rotations (R/r).
– As the coin’s center completes one revolution around the large disc’s center (path circumference 2*pi*(R+r)), the coin itself also makes one full turn due to the change in orientation as it circles the origin.
– The total number of rotations relative to a fixed direction is the sum of rotations from rolling and rotation from revolution.
– Total rotations = (R/r) + 1 = 10 + 1 = 11.