A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm .

Question

13-11-2019
Physics

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A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm .

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opudodennis opudodennis · Answer 1 · 2019-11-17T18:32:46+01:00

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

[tex]0.0192 kgm^{2}/s[/tex]

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

[tex]L = \frac{1}{{12}}m{l^2}\omega[/tex]

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for [tex]\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}[/tex]

johanrusli johanrusli · Answer 2 · 2019-12-13T05:50:42+01:00

Its angular momentum is 0.0192 kg.m²/s

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Further explanation

Complete Question:

A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm. Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second

Given:

mass of the baton = m = 0.120 kg

length of the baton = R = 80.0 cm = 0.8 m

angular velocity = ω = 3.00 rad/s

Asked:

angular momentum = L = ?

Solution:

[tex]L = I \times \omega[/tex]

[tex]L = ( \frac{1}{12} m R^2 ) \omega[/tex]

[tex]L = \frac{1}{12} m \omega R^2[/tex]

[tex]L = \frac{1}{12} \times 0.120 \times 3.00 \times 0.8^2[/tex]

[tex]L = \frac{12}{625}[/tex]

[tex]L \approx 0.0192 \texttt{ kgm}^2\texttt{/s}[/tex]

[tex]\texttt{ }[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant