IF YOU CANNOT GIVE THE ANSWER IN VECTOR FORM WHEN IT ASKS FOR IT DO NOT ANSWER T

Question

IF YOU CANNOT GIVE THE ANSWER IN VECTOR FORM WHEN IT ASKS FOR IT DO NOT ANSWER THIS QUESTION.

Two small objects each of mass m = 0.3 kg are connected by a lightweight rod of length d = 0.9 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 28 m/s and v2 = 61 m/s and are subjected to external forces whose magnitudes are F1 = 49 N and F2 = 28 N. The distance h = 0.4 m, and the distance w = 0.7 m. The system is moving in outer space. THE ANSWERS ARE ALL IN VECTOR FORM!

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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 = 40 cm

(0.4 m; the radius of rotation is 0.2 m). The barbell spins clockwise with angular speed 93 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
m/s

(b)    

of ball 1 (Enter the magnitude.)
kg · m2/s

(c)    

of ball 2 (Enter the magnitude.)
kg · m2/s

(d)    

(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 barbell
kg · m2

(i)    

expressed as a vector
rad/s

(j)    

of the barbell (Enter the magnitude.)
kg · m2/s

(k)    

Krot

J

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Calculate the angular momentum for a rotating disk, sphere, and rod. ALL THE ANSWERS ARE IN VECTOR FORM

(a) A uniform disk of mass 17 kg, thickness 0.5 m, and radius 0.7 m is located at the origin, oriented with its axis along the y axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the +y axis and look toward the origin at the disk). The disk makes one complete rotation every 0.3 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (Express your answer for rotational angular momentum in vector form.)

kg · m2/s

Krot

(b) A sphere of uniform density, with mass 21 kg and radius 0.4 m, is located at the origin and rotates around an axis parallel with the x axis. If you stand somewhere on the +x axis and look toward the origin at the sphere, the sphere spins counterclockwise. One complete revolution takes 0.8 s. What is the rotational angular momentum of the sphere? What is the rotational kinetic energy of the sphere? (Express your answer for rotational angular momentum in vector form.)

Krot

(c) A cylindrical rod of uniform density is located with its center at the origin, and its axis along the z axis. Its radius is 0.05 m, its length is 0.7 m, and its mass is 2 kg. It makes one revolution every 0.08 s. If you stand on the +x axis and look toward the origin at the rod, the rod spins clockwise. What is the rotational angular momentum of the rod? What is the rotational kinetic energy of the rod? (Express your answer for rotational angular momentum in vector form.) ALL THE ANSWERS ARE IN VECTOR FORM!!!

Krot

Lrot

kg · m2/s

Krot

Explanation / Answer

1st question

a) linear momentum ptotal of the system:

?...?.....?........?........?

P = P1 + P2 = m v1 + m v2 = (0.3 * 28) + (0.3 * 61) = 26.7 kg·m/s along positive X-axis

I.e. p = 26.7 i + 0 j +0k

b) Vcm = P / M = 26.7/ 0.6 = 44.5 m/s along positive X-axis

i.e. Vcm = 44.5 I + 0j +0k

c) total angular moment relative A:

?..?.....?......?.....?.......?...........

L = L1 + L2 = (r1 x mv1) + (r2 X mv2)

Taking into account the sin of the angles in the cross product and the geometry it results:

r1 sin?1 = d+h

r2 sin?2 = h

then L (along negative Z-axis):

Ltot = [mv1 (d+h)] + [mv2 h] = - [0.3 · 28 · (0.9+0.4)] + [0.3 · 61 · 0.4] =

= - [17.55 + 4.77] = - 18.24 kg·m²/s

i.e. Ltot = 0i + 0k – 18.24k

e) (Let's do first the translational angular momentum).

?..........?...........?

Ltrans = Rcm X MVcm

We need Rcm

?...........?.......?

Rcm = m r1 + m r2 / M

as the two masses are equal, Rcm is easy to calculate because it is in the rod center d/2 with coordinates:

Rcm = (d/2+h, w) = (0.85, 0.7) m

Then, taking into account the cross product Rcm X Vcm:

Ltrans = M Vcm · (d/2+h) = 0.6 · 44.5 · (0.85) = - 22.695 kg·m²/s along negative Z-axis.

i.e Ltrans = 0i + 0j – 22.695k

d)

?......?......?

Ltot = Lrot + Ltrans

Then:

Lrot = Ltot - Ltrans = (- 18.24) - (- 22.695) = 4.455 kg·m²/s along positive Z-axis

i.e. Lrot = 0i + 0j + 4.455k

f) F · ?t = ?P

where F is the net force:

?..?.....?

F = F1 + F2 = 49 - 28 = 21 N (along positive X-axis)

21 · 0.16 = 3.36 kg·m/s

We must add the previous P calculated in a), then

Ptot = 26.7 + 3.36 = 30.06 kg·m/s (along X-axis)

i.e Ptot = 30.06i + 0j +0k

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