A thin, cylindrical rod = 28.0 cm long with a mass m = 1.20 kg has a ball of dia

Question

A thin, cylindrical rod

= 28.0 cm

long with a mass

m = 1.20 kg

has a ball of diameter

d = 10.00 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.

(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy?
J

(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 33.0 cm?

vswing is  ---Select--- greater than less than vfall by  %.

Explanation / Answer

initial energy Ei = m*g*l/2 + M*g*(l+R)

final energy after 90 degrees rotation Ef = KErot = (1/2)*I*w^2

from energy conservation

Ef = Ei

KE = 8.1144 J

-----------------------

moment of inertia = I = (1/3)*m*l^2 + (2/5)*M*R^2 + M*(R+l)^2

I = (1/3)*1.2*0.28^2 + (2/5)*2*0.05^2 + 2*(0.28+0.05)^2

I = 0.2512 kg m^2

from energy conservation

Ef = Ei

(1/2)*0.2512*w^2 = 8.1144

w = 8.04 rad/s

=====================

(c)

vswing = (R+l)*w = (0.05+0.28)*8.04 = 2.65 m/s

============

(d)

for free fall vfall = sqrt(2*g*h) = sqrt(2*9.8*0.33) = 2.54 m/s

vswing is greater than Vfall

(Vswing-vfall)/vfall*100 = 4.3 %

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