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 kgkg and length 80.0 cmcm .Figure1 of 2Part APart completeInitially, the baton is spinning about an axis through its center at angular velocity 3.00 rad/srad/s . (Figure 1)What is the magnitude of its angular momentum about a point where the axis of rotation intersects the center of the baton?Express your answer in kilogram meters squared per second.View Available Hint(s)for Part A1.92×10−2 kg⋅m2/skg⋅m2/sSubmitPrevious Answers CorrectPart BWith a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis passing through its end at the same angular velocity 3.00 rad/srad/s as before. (Figure 2) What is the new magnitude of the angular momentum of the rod about a point where the axis of rotation intersects the end of the baton?Express your answer in kilogram meters squared per second.View Available Hint(s)
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 kgkg and length 80.0 cmcm .Figure1 of 2Part APart completeInitially, the baton is spinning about an axis through its center at angular velocity 3.00 rad/srad/s . (Figure 1)What is the magnitude of its angular momentum about a point where the axis of rotation intersects the center of the baton?Express your answer in kilogram meters squared per second.View Available Hint(s)for Part A1.92×10−2 kg⋅m2/skg⋅m2/sSubmitPrevious Answers CorrectPart BWith a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis passing through its end at the same angular velocity 3.00 rad/srad/s as before. (Figure 2) What is the new magnitude of the angular momentum of the rod about a point where the axis of rotation intersects the end of the baton?Express your answer in kilogram meters squared per second.View Available Hint(s)
Solution
Part A:
The angular momentum (L) of an object can be calculated using the formula:
L = I * ω
where I is the moment of inertia and ω is the angular velocity.
For a uniform rod rotating about its center, the moment of inertia (I) is given by:
I = (1/12) * m * l^2
where m is the mass of the rod and l is the length of the rod.
Given that m = 0.120 kg and l = 80.0 cm = 0.8 m, we can calculate I as:
I = (1/12) * 0.120 kg * (0.8 m)^2 = 0.0064 kg*m^2
Given that ω = 3.00 rad/s, we can calculate L as:
L = 0.0064 kgm^2 * 3.00 rad/s = 0.0192 kgm^2/s
Part B:
For a uniform rod rotating about its end, the moment of inertia (I) is given by:
I = (1/3) * m * l^2
We can calculate I as:
I = (1/3) * 0.120 kg * (0.8 m)^2 = 0.0256 kg*m^2
Given that ω = 3.00 rad/s, we can calculate L as:
L = 0.0256 kgm^2 * 3.00 rad/s = 0.0768 kgm^2/s
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