Question

A uniform rod of mass m and length l is struck at an end by a force F 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

Answer:

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Answer 2

(a). The speed of the center of mass is $\dfrac{Ft}{m}\ m/s$.

(b). The angular speed of the rod about the center of mass $\dfrac{6Ft}{ml}$.

(c). The kinetic energy of the rod is $2\dfrac{F^2t^2}{m}$

(d). The angular momentum of the rod about the center of mass is $\dfrac{Ftl}{2}$

Explanation:

Given that,

Mass of rod = m

Length = l

Force  = F

Time interval = t

(a). We need to calculate the acceleration

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

We need to calculate the speed of the center of mass

Using equation of motion

$v=u+at$

Put the value into the formula

$v=\dfrac{Ft}{m}\ m/s$

The speed of the center of mass is $\dfrac{Ft}{m}\ m/s$.

(b). We need to calculate the angular speed of the rod about the center of mass

Using formula of angular speed

$I\omega=r\times p$

Put the value of moment of inertia

$(\dfrac{ml^2}{12})\times\omega=\dfrac{l}{2}\times mv$

$(\dfrac{ml^2}{12})\times\omega=\dfrac{l}{2}\times Ft$

$\omega=\dfrac{6Ft}{ml}$

The angular speed of the rod about the center of mass $\dfrac{6Ft}{ml}$.

(c). We need to calculate the kinetic energy of the rod

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

Put the value into the formula

$K.E=\dfrac{1}{2}m(\dfrac{Ft}{m})^2+\dfrac{1}{2}\times(\dfrac{ml^2}{12})\times36\times\dfrac{F^2t^2}{m^2l^2}$

$K.E=2\dfrac{F^2t^2}{m}$

The kinetic energy of the rod is $2\dfrac{F^2t^2}{m}$

(d). We need to calculate the angular momentum of the rod about the center of mass

Using formula of the angular momentum

$L=mvr$

Put the value into the formula

$L=m\times\dfrac{Ft}{m}\times\dfrac{l}{2}$

$L=\dfrac{Ftl}{2}$

The angular momentum of the rod about the center of mass is $\dfrac{Ftl}{2}$

Hence, (a). The speed of the center of mass is $\dfrac{Ft}{m}\ m/s$.

(b). The angular speed of the rod about the center of mass $\dfrac{6Ft}{ml}$.

(c). The kinetic energy of the rod is $2\dfrac{F^2t^2}{m}$

(d). The angular momentum of the rod about the center of mass is $\dfrac{Ftl}{2}$

Learn more :

Topic : angular momentum

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