Question

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1). (a) At this instant, what are the magnitude and direction of its angular momentum relative to point O? (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?

Answer 1

Concepts and reason

The concepts of angular momentum and rate of change of angular momentum are required to solve the problem.

First, determine the magnitude of angular momentum by using the relation between mass, velocity, and radial distance. The direction of angular momentum is determined by using right hand rule. Then determine the rate of change of angular momentum by taking the derivative of the angular momentum expression.

Fundamentals

The angular momentum of an object of mass m relative to some point O is given by the expression,

$L = mvr\sin \theta$

Here, v is the velocity of the object, r is the distance from point O to the object, and $\theta$ is the angle between r and v.

The direction of angular momentum is determined by using right hand rule. Curl the fingers of right hand such that they form a rotation from vector r to vector v, then the thumb will give the direction of angular momentum.

The rate of change of angular momentum is equal to the torque and it is given as,

$\tau = \frac{{dL}}{{dt}}$

Torque about a point due to force F can be calculated by using tangential force method and it is given as,

$\tau = r{F_{\rm{t}}}$

Here, r is the distance from the axis of rotation to the point where force is applied and ${F_{\rm{t}}}$ is the tangential component of force.

(a)

The following figure shows a rock with velocity v moving horizontally. The distance between the point O and P is r. The angle between the vector r and vector v is $\theta$ . The angle $\phi$ is equal to ${36.9^{\rm{o}}}$ .

Refer figure 1, and determine the angle $\theta$ . The angle $\theta$ is given as,

$\begin{array}{c}\\\theta = {180^{\rm{o}}} - {36.9^{\rm{o}}}\\\\ = {143.1^{\rm{o}}}\\\end{array}$

Determine the magnitude and direction of angular momentum.

The magnitude of angular momentum is given as,

$L = mvr\sin \theta$

Here, m is the mass of the rock, v is the velocity of the rock, and $\theta$ is the angle between v and r.

Substitute 2.00 kg for m, 12.0 m/s for v, 8.00 m for r, and ${143.1^{\rm{o}}}$ for $\theta$ in the above equation.

$\begin{array}{c}\\L = \left( {2.00{\rm{ kg}}} \right)\left( {12.0{\rm{ m/s}}} \right)\left( {8.00{\rm{ m}}} \right)\sin {143.1^{\rm{o}}}\\\\ = 115{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/s}}\\\end{array}$

The direction of angular momentum is determined by using right hand rule. Curl the fingers of right hand such that they are rotating from vector r to v. Then thumb will point into the page. Thus, the direction of angular momentum is into the page.

(b)

Determine the rate of change of angular momentum.

The rate of change of angular momentum is given as,

$\tau = \frac{{dL}}{{dt}}$

According to tangential force method, torque about a point due to force F is given as,

$\tau = r{F_{\rm{t}}}$

Here, r is the distance from the axis of rotation to the point where force is applied and ${F_{\rm{t}}}$ is the tangential component of force.

Substitute $mg\cos \phi$ for ${F_{\rm{t}}}$ in equation $\tau = r{F_{\rm{t}}}$ and determine the torque.

$\tau = mgr\cos \phi$

Substitute 2.00 kg for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 8.00 m for r, and ${36.9^{\rm{o}}}$ for $\phi$ in equation $\tau = mgr\cos \phi$ and determine the rate of change of angular momentum.

$\begin{array}{c}\\\tau = \left( {2.00{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {8.00{\rm{ m}}} \right)\cos \left( {{{36.9}^{\rm{o}}}} \right)\\\\ = 125{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

The direction of torque is determined by using right hand rule. Curl the fingers of right hand such that they are rotating from vector r to force vector. Then thumb will point out of the page. Thus, the direction of rate of change of angular momentum is out of the page.

Ans: Part a

The magnitude of angular momentum is $115{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/s}}$ and direction is into the page.