Question

Two uniform thin identical rods each of mass 15 g and length 30 cm are

joined so as to form a cross as shown in figure. The system rotates about

perpendicular axis throw from its center of mass with angular velocity 20

rad/s and it is moving on circular path of radius 2 m. Calculate the total

angular momentum when its tangential velocity 10 m/

Answer 1

Answer:

The moment of inertia for each rod is I=12ml2 , about the middle point and perpendicular to the plane thus making the sum total for two rods as 2I , i.e. I=6ml2

Answer 2

Given:

Given,

$& m=15g \\ & \\ & l=30cm \\ & \\ & \omega =20rad/s \\ & \\ & r=2m \\ & \\ & {{v}_{t}}=10m/s$

To find:

We need to calculate the value of angular momentum when tangential velocity is given.

Solution:

We know that, formula of angular momentum is  

$& L=I\omega \\ & \\ & L=\frac{2M{{R}^{2}}\omega }{5}$

From the first equation, $\[I=\frac{2M{{R}^{2}}}{5}\]$ applied in the above equation.

Where, M is the mass of the rod and R is the radius of the rod and $\[\omega \]$ is the angular momentum.

Then substituting the values within the above the equation,

$& L=\frac{2M{{R}^{2}}\omega }{5} \\ & \\ & m=15g=15\times {{10}^{-3}}g \\ & \\ & L=\frac{2\times 15\times {{10}^{-3}}\times 2\times 20}{5} \\ & \\ & L=240\times {{10}^{-3}}kg\text{ }{{\text{m}}^{2}}\text{ rad/s} \\ &$

To cancel the unit radian, we have to multiply the value of  $2\pi$

$& L=240\times {{10}^{-3}}\times 2\pi \\ & \\ & \text{ = 240}\times {{10}^{-3}}\times 2\times 3.14 \\ & \\ & \text{ = 1507}\text{.2}\times {{10}^{-3}} \\ & \\ & \text{L = 1}\text{.51 kg }{{\text{m}}^{2}}/s$

Therefore, the angular momentum of rod is $\[\text{L = 1}\text{.51 kg }{{\text{m}}^{2}}/s\]$.