Question

A bullet of mass 10g and speed 500m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0m wide and weighs 12kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint. The moment of inertia of the door about the vertical axis at one end is $\dfrac{ML^2}{3}$)

Answer 1

Hint: When a bullet hits the door, it imparts the angular momentum on the door and forces it to rotate. Find the possible axis about which door can rotate. Use the law of conservation of angular momentum. By writing relation for initial and final angular momentum calculate the angular speed of the door.

Complete step-by-step answer:
Question explanation- We have a bullet of a gun whose velocity (v) is 500m/s after firing. The mass (m) of a bullet is 10 g. Now we have a door whose breadth is 1.0m and the weight of the door is 12kg. It is hinged on one side. Now to imagine this case, let's imagine our home door which is hinged at one side and we can rotate or move it about the vertical axis. The bullet hits a door at the center of the door. Now find the angular speed of the door when bullets get hit.
Let’s take the side view of the bullet hitting the door.
When a bullet hits the door, there must be impulsive forces between the door and bullet.
If we consider door and bullet as a system then, the only internal force acts i.e. torque of external is zero. When the torque gets zero, angular momentum is conserved.
By using conservation of momentum concept, we can calculate the initial and final velocity of the door. Because angular velocity is due to bullets only so at the initial stage the door is at rest and after bullets get hit it starts rotating.
Angular momentum is given by (in magnitude form),
 $ \left| {\vec{L}} \right|=mv{{r}_{\bot }} $
 $ {{r}_{\bot }}= $ Distance between a line of an axis that is the door axis and centre of the door.
If width is 1m the $ {{r}_{\bot }}=0.5m $
Therefore the angular momentum of bullet is given by,
 $ {{\left| {\vec{L}} \right|}_{i}}={{m}_{b}}{{v}_{b}}{{r}_{\bot }} $
For final angular velocity, now the bullet and door both are rotating, that is the system is purely rotational in motion.
If a system is a purely rotational motion then, angular momentum is,
 $ {{\left| {\vec{L}} \right|}_{f}}=I\omega =({{I}_{door}}+{{I}_{bullet}})\omega =(\dfrac{M{{L}^{2}}}{3}+\dfrac{M{{L}^{2}}}{4})\omega $
Given condition is an angular moment just after bullet embedded and it is given by,
\[{{\left| {\vec{L}} \right|}_{f}}=(\dfrac{12}{3}+\dfrac{{{10}^{-2}}}{4})\omega ={{\left| {\vec{L}} \right|}_{i}}={{m}_{b}}{{v}_{b}}{{r}_{\bot }}\]
\[\begin{align}
  & (\dfrac{12}{3}+\dfrac{{{10}^{-2}}}{4})\omega =\dfrac{{{10}^{-2}}\times 500}{2} \\
 & By\ Solving, \\
 & \omega =\dfrac{1000}{1601}=0.625rad/\sec \\
\end{align}\]
So the angular speed of the door is \[0.625rad/\sec \].

Note: Note that, if the question is asked to calculate angular quantity then there must be an object which does rotation. We can also calculate angular velocity by using a relation between translational velocity and angular velocity; $ v=r\omega $. In question, it is mentioned that without friction it is rotating means, so you have to neglect the frictional forces.