A coin with a diameter of 2.40 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 18.0 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 1.90 rad/s2, how far does the coin roll before coming to rest

The coin comes to rest after covering 1.023 m.

Explanation:

initial angular speed, wo = 18 rad/s

final angular speed, w = 0 rad/s

angular acceleration, = 1.9 rad/s2

radius, r = 1.2 cm =  0.012 m

initial linear velocity, u = r wo = 0.012 x 18 = 0.216 m/s

final linear speed, v = r w = 0.012 x 0 = 0 m/s

linear acceleration, a = r x angular acceleration = 0.012 x 1.9 = 0.0228 m/s2

let the distance traveled is s.

Use third equation of motion

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft