Question

A thin, cylindrical rod l = 20.8 cm long with a mass m = 1.20 kg has a ball of diameter d = 6.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge. m (a) After the combination rotates through 90 degrees, what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? rad/s (c) What is the linear speed of the center of mass of the ball? m/s (d) How does it compare with the speed had the ball fallen freely through the same distance of 23.8 cm? Vswing is greater than Vfall by

Circle answers please

Answer 1

a) rotational kinetic enetgy = loss of potential energy

= m*g*(l/2) + M*g*(l + d/2)

= 1.2*9.8*(0.208/2) + 2*9.8*(0.208 + 0.06/2)

= 5.89 J <<<<<<<<<<<------------------------Answer

b) moment of inertia of the system, I = m*l^2/3 + M*(l + d/2)^2

= 1.2*0.208^2/3 + 2*(0.208 + 0.06/2)^2

= 0.1306 kg.m^2

KE = (1/2)*I*w^2

==> w = sqrt(2*KE/I)

= sqrt(2*5.89/0.1306)

= 9.50 rad/s <<<<<<<<<<<------------------------Answer

c) v_swing = (l + d/2)*w

= (0.208 + 0.06/2)*9.5

= 2.26 m/s <<<<<<<<<<<------------------------Answer

d) v_fall = sqrt(2*g*(l + d/2))

= sqrt(2*9.8*(0.208 + 0.06/2))

= 2.16

(v_swing - v_fall)/v_swing = (2.26 - 2.16)/2.26

= 0.0442

= 4.42 % <<<<<<<<<<<------------------------Answer