Question

A uniform rod of mass m and length l is struck at an end by a force F is perpendicular to the rod for a short time interval t. Calculate
(a) The speed of the centre of mass, (b) The angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

Answer 1

The rod struck by a force F is expected to create an impulse at the rod's end and rotate the rod about another end. The angular acceleration will be determined by the force given to the rod. The center of mass will lie at the mid point of the rod.
Formula used:
Position of centre of mass is determined by:
$R_{CM} = \dfrac{\int x dm}{M}$
Moment of inertia of rod perpendicular to length at one end:
$I = \dfrac{ML^2}{3}$
Analogues to linear equations of motion, we will use:
$\omega_f = \omega_i + \alpha t$.
In a similar manner, kinetic energy for rotational motion is:
$K.E. = \dfrac{1}{2} I \omega^2$.

Complete answer:
First we determine the position of centre of mass for a rod of length L and mass M,
$R_{CM} = \dfrac{\int x dm}{M}$
here dm is the mass of small element dx and the rod has density $\rho$ which can be written as M/L. So,
$dm = \rho dx$
$R_{CM} = \dfrac{\rho}{M} \int x dx = \dfrac{\rho}{M} \dfrac{x^2}{2}$.
For the rod of length L, the limits for the integral has to go from 0 to L, so we may write:
$R_{CM} = \dfrac{\rho}{M} \dfrac{L^2}{2}$.
Keeping the value of density, we get:
$R_{CM} = \dfrac{L}{2}$.
This expression clearly indicates that the centre of mass for the rod lies at half the rod length.

(a). The velocity for centre of mass will be determined based on its distance from the axis as
$v = r \omega$.
We already know that we have to keep r = L/2. Now, we determine the angular velocity.
We know that torque can be expressed in following two ways:
1. $\tau = I \alpha$,
2. $\tau = r F$.
Therefore combining these we get the angular acceleration as:
$\alpha = \dfrac{FL}{I} = \dfrac{3F}{ML}$,
where we kept the moment of inertia of the rod about its end.

As we obtained the angular acceleration, angular velocity can be obtained as:
$\omega= \alpha t = \dfrac{3Ft}{ML}$.
Therefore, the velocity of centre of mass can now be written as:
$v = (L/2) \dfrac{3Ft}{ML} = \dfrac{3Ft}{2M}$.

(b). The angular speed will be same for every point on the road so for the centre of mass as we derived already we can write:
$\omega = \dfrac{3Ft}{ML}$.

(c). As we write for the case of linear motion, the kinetic energy for rotational motion can be written as:
$K.E. = \dfrac{1}{2} I \omega^2$.
To keep the values here we make slight changes as:
$\omega= \alpha t = \dfrac{FLt}{I}$.
So, kinetic energy expression can now be written as:
$K.E. = \dfrac{(FLt)^2}{2I} = \dfrac{3(Ft)^2}{2M}$
where we have substituted the moment of inertia, I.

(d). The angular momentum can be expressed as:
$L_r = I \omega$.
Making substitutions here, we get,
$L_r = I \times \dfrac{FLt}{I} = FLt$.

Note:
The expressions we use here can be derived easily knowing the definitions in linear form. For example, we have returned angular momentum as a product of moment of inertia and angular velocity. In rotational motion angular velocity is analogous to linear velocity and moment of inertia is analogous to mass and we know that linear momentum is mv. So, all the formulas become simple to remember if we know the analogy.