Question

If a mass of 0.8 kg is constructed at one end of a 0.5m long fiber and rotates in a horizontal circle with an angle of 10 rads, then its angular momentum about the center of rotation, in SI units

Answer 1

Explanations & Calculations

• The fiber makes the angle 10 degrees with the vertical.
• And take the speed of the particle to be $\small v$ and the tension of the fiber to be $\small T$
• Then what needs to be found is $\small mvr$.
• When a particle rotates about a point with some speed, the angular momentum is given by the above expression.
• Now, applying Newton's second law for the motion of the particle both on horizontal & verticle directions (as it rotates in a 10 degree slanted position), the speed of it can be calculated.

$\qquad\qquad \begin{aligned} \small \uparrow F&= \small ma\\ \small T\cos\theta-mg&= \small0(as \uparrow a=0)\\ \small T\cos \theta &= \small mg\cdots(1)\\ \\ \small \leftarrow F&= \small ma\\ \small T\sin\theta &=\small m\frac{v^2}{r}\cdots(2)\\ \end{aligned}\\ \\ \small\qquad \text{By (2)/(1),}\\ \qquad\qquad\qquad\qquad \begin{aligned} \small \tan\theta &= \small \frac{v^2}{rg}\\ \small v&= \small \sqrt{\tan\theta\cdot rg} \end{aligned}$

• Now, $\small r$ is not the length of the fiber but some projection of it as r should be the radial distance of the circle in which the mass rotates.

$\qquad\qquad \begin{aligned} \small r&= \small l\sin10=0.5\sin10=0.0868m \end{aligned}$

• Then the speed is

$\qquad\qquad \begin{aligned} \small v&= \small \sqrt{\tan10\times0.0868m\times 9.8ms^{-2}}\\ &= \small 0.387ms^{-1} \end{aligned}$

• Then the angular momentum is

$\qquad\qquad \begin{aligned} \small &= \small 0.8kg\times 0.387ms^{-1}\times 0.0868m\\ &= \small 0.0269kg m^2s^{-1} \end{aligned}$