Question

In Fig. 10-37, two particles, each with mass m 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d ! 5.6 cm and mass M ! 1.2 kg. The combination rotates around the rotation axis with the angular speed v ! 0.30 rad/s. Measured about O, what are the combination’s (a) rotational inertia and (b) kinetic energy

Answer 1

Answer:

(a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.

Explanation:

Given that,

Mass of each particle m= 0.85 kg

Length = 5.6 cm

Mass of each rod M= 1.2 kg

Angular speed = 0.30 rad/s

The moment of inertia of the rod  between  axis of rotation and mass  is

$I_{1}=\dfrac{Md^2}{3}$

The moment of inertia of the rod  between masses about center of mass is

$I_{cm}=\dfrac{Md^2}{12}$

Moment of inertial of the rod between masses about point O is

$I_{2}=M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}$

Moment of inertia of two masses is

$I_{m}=md^2+m(2d)^2$

(a). We need to calculate the total moment of inertia of the system

Using formula of moment of inertia

$I_{t}=I_{1}+I_{2}+I_{m}$

Put the value into the formula

$I_{t}=\dfrac{Md^2}{3}+M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}+md^2+m(2d)^2$

$I_{t}=\dfrac{Md^2}{3}+\dfrac{9Md^2}{4}+\dfrac{Md^2}{12}+md^2+4md^2$

$I_{t}=\dfrac{32Md^2}{12}+5md^2$

$I_{t}=2.67Md^2+md^2$

Put the value into the formula

$I_{t}=2.67\times1.2\times(5.6\times10^{-2})^2+5\times0.85\times(5.6\times10^{-2})^2$

$I_{t}=0.0233\ kg-m^{2}$

(b). We need to calculate the kinetic energy

Using formula of kinetic energy

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

Put the value into the formula

$K.E=\dfrac{1}{2}\times0.0233\times(0.30)^2$

$K.E=0.0011\ J$

Hence, (a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.