Question

An 8.80-cm-diameter, 300 g solid sphere is released from rest at the top of a 1.60-m-long, 18.0° incline. It rolls, without slipping, to the bottom. Part A You may want to review (Pages 315-317). What is the sphere's angular velocity at the bottom of the incline? Express your answer with the appropriate units. THÅR 3 ? | Value Units Submit Request Answer Part B What fraction of its kinetic energy is rotational? VALOR ?

Answer 1

Total energy at point A is 1.6m - Iw² + 1 2 + mg (htR) = EA I = moment of inertia of solid sphese of mass in and radius R about I= 2mR² also for rolling motion = Rw. its centre. EA - mg (htR) & vem (3 + 2) EA - mg (htR) + 7 mv2 Total energy at B in EB = Ilmulz & t I w12 + mgr - ? mulz + mgr By energy conservation A = GB - 1 ml vizu 2) = mgh. guitially it is at rest » vzo lo here g = 9,8 m/sec2

- 1.6 sin 18 m = 0.49 m N'= 2.63 m/sec v'- Rul R = 8,8 x 102 metre wl = V - 59.77 rad/sec R kinetic energy of rotationki Z mR²2 Ķ = MA 2 = x Box 16 x 2.6325 as M = mass of sphere = 3oogm=300x10 kg K = 0.83 J Total kinetic energy ķ= 1 m (v ) 10 fraction is dila s x 100 = x 110 l. = 57.1%.