Please show all work and make sure that the work can be easily to read and under

Question

Please show all work and make sure that the work can be easily to read and understand. Thank you very much.
,. A thin, cylindrical rod, e-200 em long with mas m 1.10 kg as a ball g has a ball of diameter a 4.00 em and mass Af2.00 kg attached to one end. The arrangement is originally end. The arrangement is originally free to vertical and stationary, with the ball at the top as shown. The combination is pivot about the bottom end of the rod after being given a slight nudge. 1N (a) What is the moment of inertia of the rod-and- ball system about the pivot? (4 pt (a) What is the moment of inertia of the rod-and-ball system about the pi At the instant the rod is horizontal, find (b) the rotational kinetic energy of the system and (3 pts (c) the angular speed of the rod and ball. 3 pts]

Explanation / Answer

a) moment of inertia of rod about the end point = m L^2 / 3

I1 = m l^2 / 3 = 1.10 x 0.02^2 / 3 =1.47 x 10^-4 kg m^2

moment of inertia of ball about its centre axis = 2 M (d/2)^2 / 5 = Md^2 / 10

using parallel axis theorem to find inertia of ball about pivot point,

I2 = Md^2/10 + M(l + d/2)^2

I2 = (2 x 0.04^2 / 10 ) + (2 x (0.02 + 0.02)^2 ) = 3.52 x 10^-3 kg m^2

moment of inertia of system = I1 + I2 = 3.67 x 10^-3 kg m^2

b) when rod is horizontal.

centre of mass of rod comes down height l/2.

so work done by gravity on rod = m g l /2 = 1.10 x 9.8x 0.02/2 = 0.1078 J

Workd done on ball by gravity = M g (l + d/2) = 2 x 9.8 x (0.02 + 0.04/2 )

= 2 x 9.8 x 0.04 = 0.784 J

total work done = change in KE

rotational KE = 0.784 + 0.1078 =0.892 J

c) Ke = Iw^2 /2

0.892 = ( 3.67 x 10^-3 ) w^2 /2

w = 22.05 rad/s

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