In the figure below a barbell spins around a pivot at its center at A. The barbe
ID: 1450791 • Letter: I
Question
In the figure below a barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 600 g (0.6 kg), at the ends of a very low mass rod of length d = 80 cm (0.8 m; the radius of rotation is 0.4 m). The barbell spins clockwise with angular speed 67 rad/s.
We can calculate the angular momentum and kinetic energy of this object in two different ways, by treating the object as two separate balls or as one barbell. Use the usual coordinate system, with x to the right, y toward the top of the page, and z out of the page, toward you.
I: Treat the object as two separate balls. Calculate the following quantities.
(a) The speed of ball 1 26.8 Correct: Your answer is correct. m/s
(b)Ltrans, 1, A of ball 1 (Enter the magnitude.) = ???????????kg · m2/s
(c) Ltrans, 2, A of ball 2 (Enter the magnitude.) = ???????kg · m2/s
(d) Ltot, A (Enter the magnitude.) = ??????? kg · m2/s
(e) the translational kinetic energy of ball 1= ???????? J
(f) the translational kinetic energy of ball 2 = ???????J
(g) the total kinetic energy of the barbell = ??????????J
II: Treat the object as one barbell. Calculate the following quantities.
(h) The moment of inertia I of the barbel = ????????l kg · m2
(i) , expressed as a vector = ????????? rad/s
(j) Lrot of the barbell (Enter the magnitude.) = ??????? kg · m2/s
(k) Krot =????????????? J
Explanation / Answer
a)
w = angular speed = 67 rad/s
r = radius = 0.4 m
linear speed is given as
V1 = r w = 0.4 x 67 = 26.8 m/s
b)
I1 = moment of inertia of m1 = m1 r2 = (0.6) (0.4)2 = 0.096 kgm2
rotational momentum is given as
L1 = I1 w = 0.096 x 67 = 6.432 kgm2/s
c)
I2 = moment of inertia of m2 = m2 r2 = (0.6) (0.4)2 = 0.096 kgm2
L2 = I2 w = 0.096 x 67 = 6.432 kgm2/s
d)
Ltotal = L1 + L2 = 2 x 6.432 = 12.86 kgm2/s
e)
KE1 = (0.5) I1 w2 = (0.5) (0.096) ( 67)2 = 215.5 J
f)
KE2 = (0.5) I2 w2 = (0.5) (0.096) ( 67)2 = 215.5 J
g)
KEtotal = KE1 + KE2 = 2 (215.5) = 431 J
h)
Ibarbell = 2mr2 = 2 (0.6) (0.4)2 = 0.192 kgm2