A thin, uniform, rectangular sign hangs vertically above the door of a shop. The

Question

A thin, uniform, rectangular sign hangs vertically above the door of a shop. The sign is hinged to a stationary horizontal rod along its top edge. The mass of the sign is 2.40 kg and its vertical dimension is 50.0 cm. The sign is swinging without friction, becoming a tempting target for children armed with snowballs. The maximum angular displacement of the sign is 25.0° on both sides of the vertical. At a moment when the sign is vertical and moving to the left, a snowball of mass 570 g, traveling horizontally with a velocity of 160 cm/s to the right, strikes perpendicularly the lower edge of the sign and sticks there.

(a) Calculate the angular speed of the sign immediately before the impact. _____rad/s

(b) Calculate its angular speed immediately after the impact. ____rad/s

(c) The spattered sign will swing up through what maximum angle? ____°

Explanation / Answer

First we have to calculate the moment of inertia of the rectangular slab about the rod,using prependicular axis theorem -

Iz = Ix + Iy

Icm + md2 = Ix + Iy

mh2/3 + mw2/12 = 2 Ix

if h = w

5mh2/12 = 2 Ix

Ix = 5mh2/24

Ix = [5*2.40*0.5*0.5]/24

Ix = 0.125 kg-m2

using energy conservation principle -

mgH = 0.5 mv2

g*0.5h[1-cos 25] = 0.5 v2

0.5*0.0936 = v2

v = 0.68 m/sec will be the velocity of centre of mass

v = rw

w = 0.22/0.68

w = 0.32 rad/sec

b> as we know angular momentum will remain conserved -

Icm*w = I*w'

0.125*0.32 = [0.125+0.570*0.5*0.5]w'

0.1075 = 0.2675*w'

w' = 0.15 rad/sec

v' = 0.5*0.15

v' = 0.075 m/sec

2.40*g*0.5(1-cos x) =[2.40+0.570] 0.0752

11.76(1-cos x) = 0.016

x = 3.13 degrees

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