Asolid disk of mass m1 = 9.8 kg and radius r = 0.19 m is rotating with a constant angular velocity of ? = 32 rad/s. a thin rectangular rod with mass m2 = 3 kg and length l = 2r = 0.38 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.

1)what is the initial angular momentum of the rod and disk system? kg-m2/s

2)what is the initial rotational energy of the rod and disk system? j

3)what is the final angular velocity of the disk? rad/s

4)what is the final angular momentum of the rod and disk system? kg-m2/s

5)what is the final rotational energy of the rod and disk system? j

6)the rod took t = 5.8 s to accelerate to its final angular speed with the disk. what average torque was exerted on the rod by the disk?

1) The initial angular momentum of the rod and disk system is; L_total,i = 5.664 kg•m²/s

2) The initial rotational energy of the rod and disk system is; K_total = 90.624 J

3) The final angular velocity of the disk is; ω_d,f = 26.58 rad/s

4) The final angular momentum of the rod and disk system is; L_total,f = 5.664 kg•m²/s

5) The final rotational energy of the rod and disk system is; E_total = 75.277 J

6) The average torque exerted on the rod by the disk is; τ = 0.165 N.m

We are given;

Mass of disk; m1 = 9.8 kg

Mass of rod; m2 = 3 kg

Radius of disk; r = 0.19 m

Angular velocity of disk; ω_d = 32 rad/s

Length of rod; L = 0.38 m

1) Moment of inertia of disk is;

I_disk = ½m1•r²

I_disk = ½ × 9.8 × 0.19²

I_disk = 0.177 kg.m²

Moment of inertia of rod;

I_rod = m2•L²/12

I_rod = (3 × 0.38²)/12

I_rod = 0.0361 kg.m²

Formula for angular momentum is; L = Iω

Thus, initial angular momentum of the disk and rod system is;

L_total = (0.177 × 32) + (0.0361 × 0)

L_total,i = 5.664 kg•m²/s

2) Initial rotational kinetic energy of the disk and rod system is;

K_total = ½I_disk*(ω_d)² + ½I_rod*ω²

K_total = ½(0.177 × 32²) + 0

K_total = 90.624 J

3) To get the final angular velocity of the disk, we will use the formula;

I_disk*ω_d,i = (I_disk + I_rod)ω_d,f

Plugging in the relevant values;

0.177 × 32 = (0.177 + 0.0361)ω_d,f

ω_d,f = (0.177 × 32)/(0.177 + 0.0361)

ω_d,f = 26.58 rad/s

4) From conservation of momentum, we can say that final angular momentum. Thus;

L_total,f = 5.664 kg•m²/s

5) Final total energy is;

E_total = ½(I_disk + I_rod)(ω_d,f)²

E_total = ½(0.177 + 0.0361)(26.58)²

E_total = 75.277 J

6) Average torque was exerted on the rod by the disk is;

τ = I_rod × α

Where;

α = (ω_d,f - ω)/t

α = (26.58 - 0)/5.8

α = 4.58 rad/s²

Thus;

τ = 0.0361 × 4.58

τ = 0.165 N.m

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Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:

Part a)

Initial angular momentum of the system of rod + disc is given as

here we know that

so angular momentum is given as

Part b)

Initial rotational energy is given as

Part c)

As we know that angular momentum is conserved so we will have

Part d)

Angular momentum is conserved so final angular momentum is same as initial angular momentum

Part e)

Final total energy

Part f)

Angular acceleration of the rod is given as

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .