9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) What is the cause of the increase and decrease of kinetic energy?

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

a.

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

c.

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz