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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 Fig. E10.35. (a) At this instant, what are the magnitude and direction of its angular momentum relative to point?

Answer :

Answer:

The magnitude and direction of its angular momentum relative to point is 115.28 kg m²/s.

Explanation:

Given that,

Mass = 2.00 kg

Velocity = 12.0 m/s

Suppose the distance from point O to point p and angle is 8.00 m and 36.9°.

We need to calculate the magnitude and direction of its angular momentum relative to point

Using formula of angular momentum

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

Where, r = distance from point O to point P

v = speed

[tex]\theta[/tex] = direction

[tex]L=mvr\sin(180-\alpha)[/tex]

Put the value into the formula

[tex]L=2.00\times12.0\times8.00\sin(180-36.9)[/tex]

[tex]L=115.28\ kg\ m^2/s[/tex]

The direction of angular momentum is pointing into the page.

Hence, The magnitude and direction of its angular momentum relative to point is 115.28 kg m²/s.

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