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 55.0 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 450 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.


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



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

Answer 1

Here the maximum angular displacement of the sign board is given 25 degree

mass of the sign is 2.40 kg and its vertical side is 55 cm

Now the moment of inertia of the board about the hinge point is given as

$I = \frac{mL^2}{3}$

$I = \frac{2.40\times 0.55^2}{3} = 0.242 kg m^2$

now by energy conservation

$\frac{mgL}{2}(1 - cos\theta) = \frac{I\omega^2}{2}$

$(2.40)(9.8)(0.55)(1 - cos25) = 0.242(\omega)^2$

$\omega = 2.24 rad/s$

so angular speed just before the impact is 2.24 rad/s

Part b)

now a mass of 450 g travels in opposite direction and hit at the lower end

so here we will use angular momentum conservation

$I_1\omega_1 - mvL = (I + mL^2)\omega_2$

$0.242(2.24) - (0.45\times 1.60 \times 0.55) = (0.242 + (0.45\times 0.55^2)\omega_2$

$0.542 - 0.396 = (0.378)\omega_2$

$\omega_2 = 0.386 rad/s$

Part c)

Again we can use energy conservation

$(\frac{m_1gL}{2}+ m_2gL)(1 - cos\theta) = \frac{(I_1 + I_2)\omega^2}{2}$

$(\frac{(2.40)(9.8)(0.55)}{2} + 0.450\times 9.8 \times 0.55)(1 - cos\theta) = \frac{(0.242+ 0.136)}{2}(0.386)^2$

$8.89( 1- cos\theta) = 0.028$

$cos\theta = 0.997$

$\theta = 4.56 degree$