A majorette in a parade is performing some acrobatic twirlings of her baton. Ass
ID: 1366681 • Letter: A
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 .
a) Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s . (Figure 1) What is its angular momentum?
answer:
1.92×10?2
b)
With 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/s as before. (Figure 2) What is the new angular momentum of the rod?
Express your answer in kilogram meters squared per second.
1.92×10?2
kg?m2/sExplanation / Answer
a) given , mass m = 0.12 kg
L = 80 cm
= 0.8 m
w = 3 rad/s
Moment of Inertia of batton about center, I = m*L^2/12
= 0.12*0.8^2/12
= 6.4*10^-3 kg/m^2
angular momentum, L = I*w
= 6.4*10^-3*3
= 1.92*10^-2 kg.m^2/s
b) In this case
Moment of Inertia of batton about one end, I = m*L^2/3
= 0.12*0.8^2/3
= 2.56*10^-2 kg/m^2
w = 3 rad/s
angular momentum, L = I*w
= 2.56*10^-2*3
= 7.68*10^-2 kg.m^2/s
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