A thin, cylindrical rod 25.0 cm long with a mass m d= 10.00 vertical and station

Question

A thin, cylindrical rod 25.0 cm long with a mass m d= 10.00 vertical and stationary, with the b Is free to pivot about the bottom end of the rod after being given a slight nudge. 1.20 kg cm and mass M = 2.00 kg attached to one end. The arrangement is originally all at the top as shown in the figure below. The combination 1. 71 (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 30.0 cm? Vswing is -Select Val by

Explanation / Answer

Given,

l = 25 cm ; m = 1.2 kg ; d = 10 cm ; M = 2 kg

a)From conservation of energy, PE will get converted to KE.

PE = (1.2 x 0.25 + 2 x (0.25 + 0.05)) x 9.81 = 8.83 J

Hence, KE = 8.83 J

b)We know that,

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

I = Irod + Isphere

I = m L^2/3 + 2/5M R^2 + M (R + L)^2

I = 1.2 x 0.25^2/3 + 2/5 x 2 x 0.05^2 + 2 x (0.3)^2 = 0.207 kg-m^2

KE = 1/2 I w^2 = 8.83

w = sqrt (2 KE/I) = sqrt (2 x 8.83/0.207) = 9.24 rad/s

Hence, w = 9.24 rad/s

c)v = r w = (R + L) w

v = 0.3 x 9.24 = 2.77 m/s

Hence, v = 2.77 m/s

d)v = sqrt (2 g (R + L))

v = sqrt (2 x 9.81 x 0.3) = 2.43 m/s

% = (2.77 - 2.43)/2.77 x 100 = 12.27 %

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