Question

The figure shows an overhead view of a 2.10 kg plastic rod of length 1.20 m on a table. One end of the rod is attached to the table, and the rod is free to pivot about this point without friction. A disk of mass 55.0 g slides toward the opposite end of the rod with an initial velocity of 26.5 m/s. The disk strikes the rod and sticks to it. After the collision, the rod rotates about the pivot point.
(a)What is the angular velocity, in rad/s, of the two after the collision?
(b)What is the kinetic energy, in joules, before and after the collision?
KEi
=
J
KEf
=
J

Answer 1

(a) The angular velocity of the system after the collision is 1.61 rad/s.

(b) The kinetic energy before collision is 19.31 J and the kinetic energy after collision is 1.45 J.

Conservation of angular momentum

The angular velocity of the system is determined by applying the principle of conservation of angular momentum.

Li = Lf

P x r = Ifωf

mv x r = Ifωf

$\omega _f = \frac{mvr}{I_f} \\\\\omega _f = \frac{mvr}{\frac{1}{3}Mr^2 + mr^2 } \\\\\omega _f = \frac{mvr}{r^2(\frac{1}{3}M + m )}\\\\\omega _f = \frac{mv}{r(\frac{1}{3}M + m )}\\\\\omega _f = \frac{0.055 \times 26.5}{1.2(\frac{2.1}{3} \+ \ 0.055)} \\\\\omega_f = 1.61 \ rad/s$

Kinetic energy before the collision

Kinetic energy before the collision is calculated as follows;

$K.E_i = K.E_{ball} + K.E _{rod}\\\\K.E_i = K.E_{ball} + 0\\\\K.E_i = \frac{1}{2} \times m \times v^2\\\\K.E_i = \frac{1}{2} \times 0.055 \times 26.5^2\\\\K.E_i = 19.31 \ J$

Kinetic energy after the collision

The kinetic energy after the collision is calculated as follows;

$K.E_f =\frac{1}{2} I_f \omega_ f^2\\\\K.E_f = \frac{1}{2} (\frac{1}{3}Mr^2 + mr^2)\omega _f^2\\\\K.E_f =\frac{1}{2} [\frac{1}{3}(2.1)(1.2)^2 + (0.055)(1.2)^2]\times (1.61)^2\\\\K.E_f = \frac{1}{2} (1.008 + 0.114) \times (1.61)^2\\\\K.E_f = 1.45 \ J$

Learn more about rotational kinetic energy here: https://brainly.com/question/25803184