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A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure.

(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?

♦ Relevant knowledge

If there is a movement of the body and the body is moved from one place to another, the body has to have velocity. In all cases, the velocity will be linear.

Concepts and ReasonTo solve this problem, the concepts of angular motion and rate of change in angular momentum will be required.

The first step is to determine the magnitude and velocity of angular momentum using the relationship between mass, velocity, and radial distance. Right-hand rule is used to determine the direction of angular motion. Next, determine the rate at which angular momentum changes by using the right-hand rule.

FundamentalsThe expression gives the angular momentum relative to O of an object of mass.

[katex]L = mvr\sin \theta[/katex]

Here, v denotes the velocity of the object and r the distance from O to the object.

Right-hand rule determines the direction of angular momentum. The right-hand rule determines the direction of angular momentum by bending the fingers of the right hand so that the rotation of vector

rto vector is formed.The rate at which angular momentum changes is equal to torque. It is described as:

[katex]\tau = \frac{{dL}}{{dt}}[/katex]

The tangential force method can calculate torque around a point due to force

F. It is shown as[katex]\tau = r{F_{\rm{t}}}[/katex]

Here, r is the distance between the axis rotation and the point at which force is applied. [katex]{F_{\rm{t}}}[/katex]

(a)

The following illustration shows a horizontally moving rock with velocity V. The distance between O and P is r.

Refer to figure 1 and calculate the angle [katex]\theta[/katex].

[katex]\begin{array}{c}\\\theta = {180^{\rm{o}}} – {36.9^{\rm{o}}}\\\\ = {143.1^{\rm{o}}}\\\end{array}[/katex]

Find the direction and magnitude of angular momentum.

This is the magnitude of angular momentum.

[katex]L = mvr\sin \theta[/katex]

Here, m is a measure of the rock’s mass, v is its velocity, and [katex]\theta[/katex] is the equation.

Substitute the value:

[katex]\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}[/katex]

The right-hand rule determines the direction of angular momentum. Right-hand rule: Curl your right-hand fingers so that they rotate from vector

rinto_v. Your thumb should point in the direction of the page. The direction of angular momentum will be into the page.(b)

Calculate the rate at which angular momentum changes.

The rate at which angular momentum changes is described as

[katex]\tau = \frac{{dL}}{{dt}}[/katex]

The tangential force method calculates torque around a point caused by force

F. It is shown as:[katex]\tau = r{F_{\rm{t}}}[/katex]

Here, r is the distance between the axis rotation and the point at which force is applied. [katex]{F_{\rm{t}}}[/katex]

Substitute

mgcosϕfor [katex]{F_{\rm{t}}}[/katex] in equation [katex]\tau = r{F_{\rm{t}}}[/katex] and determine torque.[katex]\tau = mgr\cos \phi[/katex]

Substitute 2.00kg for m [katex]9.8{\rm{ m/}}{{\rm{s}}^2}[/katex] for g, 8.00 m for r, and 36.9

^{o}forϕin equation [katex]\tau = mgr\cos \phi[/katex] and calculate the change in the velocity of angular motion.[katex]\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}[/katex]

Right hand rule determines the direction of torque. Right hand rule: Curl your right-hand fingers so that they rotate from vector

to vector. Next, point your thumb out from the page. The direction of the rate of change in angular momentum is thus out of the page.Ans:a. The magnitude of angular momentum in [katex]115{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/s}}[/katex] and direction is into the page.

b. The magnitude of angular momentum in [katex]125{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/s}}[/katex] and direction is out of the page.