Question

016 10.0 points A regulation basketball has a 31 cm diameter and may be approximated as a thin spherical shell. How long will it take a basketball starting from rest to roll without slipping 4.0 m down an incline that makes an angle of 31.4° with the horizontal? The acceleration of gravity is 9.81 m/s2 Answer in units of s.

Answer 1

D = 31 cm theta = 31.4 g = 9.81 m/s^2 s = 15.5 times 10^- 2 m s = 4.0 m mgh = 1/2 mv^2 + 1/2 I w^2 mgh = 1/2 mv^2 + 1/2 m k^2 v^2/r^2 g h = v^2/2 + k^2 v^2/2r^2 rightarrow v^2/2 (1 + k^2/r^2) = g h v^2 = 2 g h/(1 + k^2/r^2) = 2 g h/(1 + 2/3) but k = squareroot i/m = squareroot 2/3 m s^2/m k = squareroot 2/3 s v^2 = 2gh/5 times 3 v^2 = 29.5 sin theta times 3/5 Therefore t = v/g sin theta = squareroot 2g s sin theta 3/5/g sin theta t = squareroot 615/g sin theta = 0.486 second t = 0.486 second (Because g sin theta) v = u + at v = at