- A Barbell Spins Around A Pivot At Its Center At A The Barbell Consists Of Two Small Balls Each With Mass 550 Grams 0 1 (73.43 KiB) Viewed 47 times
object in two different ways, by treating the object as two
separate balls, or as one barbell.
I: Treat the object as two separate balls
(a) What is the speed of ball 1?
|| = m/s
(b) Calculate the translational angular
momentum trans, 1, A of
just one of the balls (ball 1).
|trans, 1, A|
= kg · m2/s
zero magnitude; no direction out of
page into page
(c) Calculate the translational angular
momentum trans, 2, A of the
other ball (ball 2).
|trans, 2, A|
= kg · m2/s
into page out of
page zero magnitude; no direction
(d) By adding the translational angular momentum of ball 1 and the
translational angular momentum of ball 2, calculate the total
angular momentum of the barbell, tot, A.
|tot, A| = kg
· m2/s
zero magnitude; no direction into
page out of page
(e) Calculate the translational kinetic energy
of ball 1.
Ktrans,1 =
m||2
= J
(f) Calculate the translational kinetic energy
of ball 2.
Ktrans,2 =
m||2
= J
(g) By adding the translational kinetic energy of ball 1 and the
translational kinetic energy of ball 2, calculate the total kinetic
energy of the barbell.
Ktotal = J
II: Treat the object as one barbell
(h) Calculate the moment of inertia I of the
barbell.
I = kg · m2
(i) What is the direction of the angular velocity
vector ?
into page out of
page zero magnitude; no direction
(j) Use the moment of inertia I and the angular
speed || = 80 rad/s to calculate the
rotational angular momentum of the barbell:
|rot|
= I || = kg · m2/s
zero magnitude; no direction into
page out of page
(k) How does this value, |rot|, compare to the angular momentum
|tot, A| calculated
earlier by adding the translational angular momenta of the two
balls?
|rot| < |tot, A||rot| > |tot, A| |rot| = |tot, A|
(l) Use the moment of inertia I and the angular
speed || = 80 rad/s to calculate the
rotational kinetic energy of the barbell:
Krot =
I𝜔2
= J
(m) How does this value, Krot, compare to
the kinetic energy Ktotal calculated
earlier by adding the translational kinetic energies of the two
balls?
Krot > KtotalKrot = Ktotal Krot < Ktotal
A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 550 grams (0.55 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 80 radians/s. V 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. I: Treat the object as two separate balls (a) What is the speed of ball 1? M = m/s (b) Calculate the translational angular momentum ftrans, 1, A of just one of the balls (ball 1). trans, 1, Al = kg Å m²/s zero magnitude; no direction out of page Ⓒinto page (c) Calculate the translational angular momentum Itrans, 2, A of the other ball (ball 2). I trans, 2, Al = kg Å- m²/s into page out of page zero magnitude; no direction (d) By adding the translational angular momentum of ball 1 and the translational angular momentum of ball 2, calculate the total angular momentum of the barbell, tot, A. Itot, Al- kg Å+ m²/s zero magnitude; no direction Ⓒinto page Ⓒout of page (e) Calculate the translational kinetic energy of ball 1. Ktrans,1 m2 = = J (f) Calculate the translational kinetic energy of ball 2. Ktrans,2 m2. = J (g) By adding the translational kinetic energy of ball 1 and the translational kinetic energy of ball 2, calculate the total kinetic energy of the barbell. Ktotal- II: Treat the object as one barbell (h) Calculate the moment of inertia I of the barbell. J- kg Å. m² (1) What is the direction of the angular velocity vector ? Ⓒinto page Ⓒ out of page zero magnitude; no direction
(j) Use the moment of inertia I and the angular speed || = 80 rad/s to calculate the rotational angular momentum of the barbell: |Irot| I || = kg Â. m²/s zero magnitude; no direction into page O out of page (k) How does this value, Irotl, compare to the angular momentum | Itot, Al calculated earlier by adding the translational angular momenta of the two balls? Erotitot, Al oirot > Itot, Al o rotl=Itot, Al (I) Use the moment of inertia I and the angular speed || = 80 rad/s to calculate the rotational kinetic energy of the barbell: Krot = 1w² = J (m) How does this value, Krot, compare to the kinetic energy Ktotal calculated earlier by adding the translational kinetic energies of the two balls? o Krot > Ktotal o Krot = Ktotal Krot < Ktotal