Question

A ball of mass M = 1.40 kg and radius r = 4.70 cm is attached to one end of a thin, cylindrical rod of length L = 12.0 cm and mass m = 0.880 kg. The ball and rod, initially at rest in a vertical position and free to rotate around the axis shown in the figure below, are nudged into motion. M m (a) What is the rotational kinetic energy of the system when the ball and rod reach a horizontal position? Enter a number. ffers from the correct answer by more than 10%. Double check your calculations. J (b) What is the angular speed of the ball and rod when they reach a horizontal position? rad/s (c) What is the linear speed of the center of mass of the ball when the ball and rod reach a horizontal position? m/s (d) What is the ratio of the speed found in part (C) to the speed of a ball that falls freely through the same distance? Vc) Vfreefall =

Answer 1

M с A (a) we B In the given want to find the rotational kinetic energy when the ball and rod system reach horizontal position. ven initial position, total energy = Ei U; į ki Iki= initial Kinetic energy Uia initial potential energy = =0

Ui= mg Irod L ie mg I red + mg sponere mg (2) + Mg (2+2) [ center of mass of rod 늘 By conservation of energy, we can write, final kinetic energy rotational Kinetic energy - Ei avi [ as in the final position (horizontal position) Potential total energy rotational Kinetic energy] Required rotational Kinetic energy akt = mg. + 2 + Mg (2+4) Flore 1:40*9.81 (4.70 +12 [4r = 2.81 Joule =4.7cm] ( energy zo and energy is X 0.88 x 9.81 x 12x10-2 2² ) + 2 10-2 ע Answer kg = 2.81] 71 = Ź I w2 Rotational Kinetic energy = kg The system rotates about point A. I = moment of inertia of the system ... about A - Irod + Isphere. + m2 ² + [ 2 mine + M (P+h)? I we we have considered the ball is a solid sphere of radius r= 4.7 cm]

M(+2)? + I at m ² + 2 / Mr²+ 1 x 0.88-28.122 x 1:40x (0.047)2 + 140 (4.70 +12)x10-219 1oI** 0.88 * (12)* + ? x 1.4* (4.70)2 + 1.40x (16.703 Troy q r 445-05) kg-m? = ban 4 = ² 1 0² 2.81 10-4 x 445.05* w - 2.81 AN w? 2.81x2 4u5.05 Loy 슈 Ž.ws 11:24 rad's to24 Angular speed rad| (when they reach horizontal position) The system system rotates about point A. Linear relocity of point D = N = Rw (6P+1) = (4.70+12) where R = .AD = cm = 16.70cm vo R.w 11:24 (16.70x 10-2 ) mls X 11877 m/s of center of mass Linear velo velocity of (Answer) the ball > 11871 m/s

2 ) If a ball falls freely through the same distance R = 16.70cm, v=u²+2ts initial velocity u. acceleration on ta tag = 9.81 mis? 16.170 x 102m ☺ distance s ... v²= ot S zg - Va Jags 2x9.81* 0.167 = 1.81m V (c) 7.877 - 1.04 freefall 1.81 Answer = Nes free fall ~ lou
The ball is considered as a solid sphere.

In part (d), it is not clearly mentioned what 'the same distance' actually indicates. We have considered here that a ball falls freely through the distance (L+d/2). If the 'same distance' indicates L, then velocity of the ball while it falls freely through L distance, =v = 1.53 m/s and the ratio will be approximately 1.23 then.