(a) What is the magnitude of the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 79.0 rev/min in 4.40 s

An angular speed of 79.0 rev/min = 4740 rev/s corresponds to a tangential speed of

(4740 rev/s) • (2π (13.0 in) / rev) = 123,240π in/s ≈ 387,000 in/s

and thus a tangential acceleration of

(123,240π in/s - 0) / (4.40 s) = 28,009.1π in/s² ≈ 88,000 in/s²

the force required to move the cart is divided by three or 33.33 n.

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