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

a)From conservation of energy, the intial PE of the system gets convetred to the KE when the system rotates through 90 deg

PEi = [1.2 x 0.28 + 2 x (0.05 + 0.28)] 9.8 = 9.76 J

Hence, PEi = 9.76 J

b)KE = 1/2 I w^2

I = Irod + I sphere

I = ml^2/3 + 2/5 Mr^2 + M(R + l)^2

I = 1.2 x 0.28^2/3 + 2/5 x 2 x 0.05^2 + 2 (0.05 + 0.28)^2 = 0.25 kg-m^2

1/2 x 0.25 x w^2 = 9.76 => w = 8.84 rad/s

Hence, w = 8.84 rad/s

c)v = r w

v = (0.05 + 0.28) x 8.84 = 2.92 m/s

Hence, v = 2.92 m/s

d)free fall velocity is

v = sqrt(2 g h)

v = sqrt (2 x 9.8 x 0.33) = 2.54 m/s

%age = (2.92-2.54)/2.54 x 100 = 14.96 %

Hence, v swing is greater than v fall by 14.96 %

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