Question

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

Answer 1

Answer:

(a) The "angular speed" is 5.88 rad/s.

Explanation:

Given values,

The length of the bar is L = 2m

The weight of the bar is w = 90 N

The metal bar is hanging vertically from the ceiling by a frictionless pivot  

The mass of the ball is m = 3kg

The distance between the ceiling and the ball is d = 1.5m

$\text { The "initial speed of the ball" is } V_{i}=10 \mathrm{m} / \mathrm{s}$

$\text { The "final speed of the ball" is } V_{f}=6 \mathrm{m} / \mathrm{s}$

(a) Calculating the angular speed:    

$W_{\mathrm{f}}=\left(3 \mathrm{mg} \mathrm{d} \frac{V_{i}+V_{f}}{w l^{2}}\right)$

$W_{f}=\left(3 \times 3 \times 9.8 \times 1.5 \times \frac{10+6}{90 \times 2^{2}}\right)$

$W_{f}=\frac{2116.8}{360}$

$\mathrm{W}_{\mathrm{f}}=5.88 \mathrm{rad} / \mathrm{s}$

The angular speed is 5.88 rad/s.

(b) The "angular momentum" is conserved because the torque is not exerted by "the pivot" on the system about the "axis of rotation" but the "linear momentum" is not conserved because "the pivot" exerts a "vertical" and a "horizontal force" on the system during the collision.