Physics

1. A parallel-plate capacitor has initial capacity C, the permittivity of free space is ε_o, plate area is A, the distance between plates is d. If the plate area increased by 4 times, the distance between plates becomes 2d and the permittivity of free space is 5ε_o, what is the final capacity of the parallel-plate capacitor.

Known :

The capacity of capacitor = C

The permittivity of free space = ε_o

Plate area = A

The distance between plates = d

Wanted: Capacity of the capacitor (C)

Solution :

2. The capacitor that has the largest capacity based on the graphic below is…

Solution :

The formula of the parallel-plate capacitor :

The capacity of capacitors :

The capacitor that has the largest capacity is capacitor C_3.

3.

Consider the following factors!
(1) Dielectric constant
(2) The potential difference between the plates
(3) Plate thick
(4) The surface area of the plate
(5) The distance between the plates
(6) Amount of electric charge
Factors affecting the capacities of parallel plate capacitors are …

Solution :

(1), (4) and (5)

4. The capacity comparison of the capacitor 1 and 2 is…

Known :

Capacitor 1 :

Surface area = 2A

The distance between plates = d₁

Capacitor 2 :

Surface area = A

The distance between plates = 2 d₁

Wanted: The capacity comparison of the capacitor 1 and 2

Solution :

5.

Determine comparison of the capacity of Parallel-plate capacitor I and II.

Solution :

Comparison of the capacity of Parallel-plate capacitor I and II :

6. Two parallel-plate capacitors shown in the figure below.

If A₁ = ½ A₂ and d₂ = 3 d₁ then determine the ratio of capacities of the parallel-plate capacitor between the image 2 and the image 1.

Known :

Parallel-plate capacitor I :

A₁ = 1

d₁ = 1

Parallel-plate capacitor II :

A₂ = 2

d₂ = 3

Wanted: the ratio of capacities of the parallel-plate capacitor between the image 2 and the image 1

Solution :

Parallel-plate capacitor I :

Parallel-plate capacitor II :

The ratio of capacities of the parallel-plate capacitor II and I :

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1. Determine the electric potential at a point located at 1 cm from a charge 5.0 μC. Coulomb constant (k) = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C.

Known :

The distance from charge (r) = 1 cm = 1/100 m = 0.01 m = 10^-2 m

Charge (q) = 5.0 μC = 5.0 x 10^-6 C

Coulomb constant (k) = 9 x 10⁹ Nm²C⁻²

Wanted : Electric potential (V)

Solution :

Electric potential :

Electric potential is 4.5 x 10⁶ Volt

2. Charge Q₁ = 5.0 μC and charge Q₂ = 6.0 μC. Coulomb’s constant (k) = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C. Point A located between the charges. Determine electric potential at point A.

Known :

Charge Q₁ = -5.0 μC = -5.0 x 10^-6 C

The distance of point A from Q₁ = 10 cm = 0.1 m = 10^-1 m

Charge Q₂ = 6.0 μC = 6.0 x 10^-6 C

The distance of point a from Q₂ = 10 cm = 0.1 m = 10^-1 m

Coulomb’s constant (k) = 9 x 10⁹ Nm²C⁻²

Wanted : Electric potential at point A

Solution :

Electric potential 1 :

Electric potential 2 :

Electric potential at point A :

V = V₂ – V₁

V = (54 – 45) x 10⁴

V = 9 x 10⁴

3. Charge q₁ = 5.0 μC and charge q₂ = 6.0 μC. Coulomb’s constant (k) = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C. Determine the electric potential at point A.

Known :

Charge Q₁ = -5.0 μC = -5.0 x 10^-6 C

The distance of point A from Q₁ = 40 cm = 0.4 m = 4 x 10^-1 m

Charge Q₂ = 6.0 μC = 6.0 x 10^-6 C

The distance of point A from Q₂ = 50 cm = 0.5 m = 5 x 10^-1 m

Coulomb’s constant (k) = 9 x 10⁹ Nm²C⁻²

Wanted : Electric potential at point A

Solution :

Electric potential 1 :

Electric potential 2 :

Electric potential at point A :

V = V₁ + V₂

V = (-11.25 + 10.8) x 10⁴

V = -0.45 x 10⁴

V = -4.5 x 10³

Electric potential at point A is -4.5 x 10³ Volt

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1. An electron is accelerated from rest through a potential difference 12 V. What is the change in electric potential energy of the electron?

Known :

The charge on an electron (e) = -1.60 x 10^-19 Coulomb

Electric potential = voltage (V) = 12 Volt

Wanted: The change in electric potential energy of the electron (ΔPE)

Solution :

ΔPE = q V = (-1.60 x 10^-19 C)(12 V) = -19.2 x 10^-19 Joule

The minus sign indicates that the potential energy decreases.

2. Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate.

Known :

The magnitude of the electric field between the plates (E) = 500 Volt/meter

The distance between the plates (s) = 2 cm = 0,02 m

The charge on an proton = +1.60 x 10^-19 Coulomb

Wanted : The change in electric potential energy (ΔPE)

Solution :

Electric potential :

V = E s

V = (500 Volt/m)(0.02 m)

V = 10 Volt

The change in electric potential energy :

ΔPE = q V

ΔPE = (1,60 x 10^-19 C)(10 V)

ΔPE = 16 x 10^-19 Joule

ΔPE = 1.6 x 10^-18 Joule

3. Two point charges are separated by a distance of 10 cm. Charge on point A =+9 μC and charge on point B = -4 μC. k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

Known :

Charge A (q₁) = +9 μC = +9 x 10⁻⁶ C

Charge B (q₁) = -4 μC = -4 x 10⁻⁶ C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge A and B (r) = 10 cm = 0.1 m = 10^-1 m

Wanted : The change in electric potential energy (ΔEP)

Solution :

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1. A uniform electric field E = 8000 N/C passing through a flat square area A = 10 m². Determine the electric flux.

Known :

The magnitude of the electric field (E) = 8000 N/C

Area (A) = 10 m²

θ = 0^o (the angle between the electric field direction and a line drawn a perpendicular to the area)

Wanted: Electric flux (Φ)

Solution :

The formula of electric flux :

Φ = E A cos q

Φ = electric flux (Nm²/C), E = electric field (N/C), A = area (m²), q = angle between electric field line with the normal line.

Electric flux :

Φ = E A cos q = (8000)(10)(cos 0) = (8000)(10)(1) = 80,000 = 8 x 10⁴ Nm²/C

2. A uniform electric field E = 5000 N/C passing through a flat square area A = 2 m². Determine the electric flux.

Known :

Electric field (E) = 5000 N/C

Area (A) = 2 m²

θ = 60^o (the angle between the electric field direction and a line drawn perpendicular to the area)

Wanted : Electric flux (Φ)

Solution :

Electric flux :

Φ = E A cos q = (5000)(2)(cos 60) = (5000)(2)(0.5) = 5000 = 5 x 10³ Nm²/C

3. A solid ball with 0.5 meters radius has 10 μC electric charge in its center. Determine the electrical flux pass through the solid ball.

Known :

Radius of ball (r) = 0.5 m

Electric charge (Q) = 10 μC = 10 x 10^-6 C

Wanted : Electric flux (Φ)

Solution :

Electric field :

E = k q/r²

E = (9 x 10⁹ Nm²/C²)(10 x 10^-6 C) / 0.5²

E = (90 x 10³) / 0,25

E = 360 x 10³

E = 3.60 x 10⁵ N/C

Surface area :

A = 4 π r² = 4 (3.14)(0.5)² = (12.56)(0.25) = 3.14 m²

Electric flux :

The electric field lines perpendicular to area, so that the angle between the electric field direction and a line drawn perpendicular to the area, is 0^o.

Φ = E A cos q

Φ = (3.60 x 10⁵)(3.14)(cos 0)

Φ = (11.304 x 10⁵)(1)

Φ = 11.304 x 10⁵

Φ = 1.13 x 10⁶ Nm²/C

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1. Calculate the magnitude and direction of the electric field at a point A located at 5 cm from a point charge Q = +10 μC. k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C)

Known :

Electric charge (Q) = +10 μC = +10 x 10^-6 C

The distance between point A and point charge Q (r_A) = 5 cm = 0.05 m = 5 x 10^-2 m

k = 9 x 10⁹ Nm²C⁻²

Wanted: The magnitude and direction of the electric field at point A

Solution :

The direction of the electric field at point A :

The electric charge is positive hence the direction of the electric field away from the electrical charge and points A.

2. Calculate the magnitude and direction of the electric field at a point P located at 10 cm from a point charge Q = -20 μC. k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C.

Known :

Electric charge (q) = -20 μC = -20 x 10^-6 C

The distance between point P and electric charge (r_P) = 10 cm = 0.1 m = 1 x 10^-1 m

k = 9 x 10⁹ Nm²C⁻²

Wanted: The magnitude and direction of the electric field at point P

Solution :

The direction of electric field at point A :

The electric charge is negative hence the direction of the electric field to the electrical charge.

3. Two point charges are separated by a distance of 40 cm. What is the magnitude and direction of the electric field at the point P between the two charges, that is 20 cm from point A?

Known :

Charge A (q_A) = -2 μC = -2 x 10^-6 C

Charge B (q_B) = +4 μC = +4 x 10^-6 C

The distance between charge A and point P (r_AP) = 20 cm = 0.2 m = 2 x 10^-1 m

The distance between charge B and point P (r_BP) = 20 cm = 0.2 m = 2 x 10^-1 m

Wanted : The magnitude and direction of electric field at point P.

Solution :

Charge A is negative so that the direction of the electric field points toward Q_A (to the left).

Charge B is positive so that the direction of the electric field points away from Q_B (to the left).

The total electric field at point A :

E = E_A + E_B

E = (4.5 x 10⁵) + (9 x 10⁵)

E = 13.5 x 10⁵ N/C

The direction of the electric field points toward Q_A (to the left).

4. The magnitude of the electric field is zero at…

The charge A is positive and the charge B is positive so that the magnitude of the electric field is zero located at point P, between both charges.

Known :

Charge A (q_A) = +20 μC = +20 x 10⁻⁶ C

Charge B (q_B) = +40 μC = +40 x 10⁻⁶ C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge A and the charge B = 20 cm

The charge between charge A and point P (r_AP) = a

The distance between charge B and point P (r_BP) = 20 – a

Wanted : The magnitude of the electric field is zero located at….

Solution :

The magnitude of the electric field produced by charge A at point P

Charge A is positive so that the direction of the electric field points away from charge A (to the right).

The magnitude of the electric field produced by charge B at point P :

Charge B is positive so that the direction of the electric field points away from charge B (to the left).

The total electric field at point P = 0 :

We use the quadratic formula to determine a.

The magnitude of the electric field is zero located at 8 cm from charge A or 12 cm from charge B.

5. Based on the figure below, where is the point P so that the electric field at point P is zero? (k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C)

Solution

To calculate the electric field strength at point P, assumed at point P there is a positive test charge. Q₁ is positive and Q₂ is negative, therefore point P must be on the right of Q₂ or left of Q₁. If point P is to the left of Q₁; the electric field generated by Q₁ at the point P is to the left (away from Q₁) and the electric field generated by Q₂ at the point P to the right (towards Q₁). The direction of the electric field is opposite so that the two electric fields eliminate each other so that the electric field strength at point P is zero.

Known :

Q₁ = +9 μC = +9 x 10⁻⁶ C

Q₂ = -4 μC = -4 x 10⁻⁶ C

k = 9 x 10⁹ Nm²C⁻²

Distance between charge 1 and charge 2 = 3 cm

Distance between Q₁ and point P (r_1P) = a

Distance between Q₂ and point P (r_2P) = 3 + a

Wanted : location of point P so that the electric field at point P is zero

Solution :

Point P is on the left of Q₁.

The electric field produced by Q₁ at point P:

The test charge is positive and Q₁ is positive so that the direction of an electric field to leftward.

The electric field produced by Q₂ at point P:

The test charge is positive and Q₂ is negative so that the direction of an electric field to rightward.

Net electric field at point A :

Use the quadratic formula to determine a :

a = -1.25, b = -13.5, c = -20.25

Distance between Q₂ and point P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm.

Point P is on the 1.2 cm right of Q₁.

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1. Two point charges, Q_A = +8 μC and Q_B = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 10⁹ Nm²C⁻² = 9 x 10⁹ Nm²C⁻².

Known :

Charge A (q_A) = +8 μC = +8 x 10^-6 C

Charge B (q_B) = ­-5 μC = -5 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge A and B (r_AB) = 10 cm = 0.1 m

Wanted : The magnitude of the electric force

Solution :

Formula of Coulomb’s law :

The magnitude of the electric force :

2. Two charged particles as shown in figure below. Q_P = +10 μC and Q_q = +20 μC are separated by a distance r = 10 cm. What is the magnitude of the electrostatic force.

Known :

Charge P (Q_P) = +10 μC = +10 x 10^-6 C

Charge Q (Q_Q) = +20 μC = +20 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge P and Q (r_PQ) = 12 cm = 0.12 m = 12 x 10^-2 m

Wanted : The magnitude of the electric force

Solution :

3. Three charged particles are arranged in a line as shown in figure below. Charge A = -5 μC, charge B = +10 μC and charge C = -12 μC. Calculate the net electrostatic force on particle B due to the other two charges.

Known :

Charge A (q_A) = -5 μC = -5 x 10^-6 C

Charge B (q_B) = +10 μC = +10 x 10^-6 C

Charge C (q_C) = -12 μC = -12 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between particle A and B (r_AB) = 6 cm = 0.06 m = 6 x 10^-2 m

The distance between particle B and C (r_BC) = 4 cm = 0.04 m = 4 x 10^-2 m

Wanted : The magnitude and the direction of net electrostatic force on particle B

Solution :

The net force on particle B is the vector sum of the force F_BA exerted on particle B by particle A and the force F_BC exerted on particle B by particle C.

The force F_BA exerted on particle B by particle A :

 

The direction of the electrostatic force points to particle A (point to left).

The force F_BC exerted on particle B by particle A :

 

The direction of the electrostatic force points to particle C (point to right).

The net electrostatic force on particle B :

F_B = F_AB – F_BC = 675 N – 125 N = 550 Newton.

The direction of the net electrostatic force on particle B points to particle C (points to the right).

4. +Q₁ = 10 μC, +Q₂ = 50 μC and Q₃ are separated as shown in the figure below. What is the electrostatic charge on particle 3 if the net electrostatic force on particle 2 is zero.

Known :

Charge 1 (q₁) = +10 μC = +10 x 10^-6 C

Charge 2 (q₂) = +50 μC = +50 x 10^-6 C

The distance between charge 1 and 2 (r₁₂) = 2 cm = 0.02 m = 2 x 10^-2 m

The distance between charge 2 and charge 3 (r₂₃) = 6 cm = 0.06 m = 6 x 10^-2 m

The net electrostatic force on particle 2 (F₂) = 0

Wanted : charge 3 (q₃)

Solution :

The net force on particle 2 is the vector sum of the force F₂₁ exerted on particle 2 by particle 1 and the force F₂₃ exerted on particle 2 by particle 3.

The force F₂₁ exerted on particle 2 by particle 1 :

The direction of the electrostatic force points to particle 3 (point to right).

The force F₂₃ exerted on particle 2 by particle 3 : 

The direction of the electrostatic force points to particle 1 (point to left).

The net electrostatic force on particle 2 = 0 :

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1. An object with the moment of inertia of 2 kg m² rotates at 1 rad/s. What is the angular momentum of the object?

Known :

Moment of inertia (I) = 2 kg m²

Angular speed (ω) = 1 rad/s

Wanted : Angular momentum (L)

Solution :

Formula of angular momentum :

L = I ω

L = angular momentum (kg m²/s), I = moment of inertia (kg m²), ω = angular speed (rad/s)

The angular momentum :

L = I ω = (2)(1) = 2 kg m²/s

2. A 2-kg cylinder pulley with radius of 0.1 m rotates at a constant angular speed of 2 rad/s. What is the angular momentum of the pulley ?

Known :

Mass of pulley (m) = 2 kg

Radius of pulley (r) = 0.1 m

Angular speed (ω) = 2 rad/s

Wanted : Angular momentum

Solution :

Formula of moment of inertia for solid cylinder :

I = 1/2 m r²

I = moment of inertia (kg m²), m = mass (kg), r = radius (m)

Moment of inertia :

I = 1/2 (2)(0.1)² = (1)(0.01) = 0.01 kg m²

The angular speed :

L = I ω = (0.01)(2) = 0.02 kg m²/s

3. A 2-kg uniform sphere with radius of 0.2 m rotates at 4 rad/s. What is the angular momentum of the ball.

Known :

Mass of ball (m) = 2 kg

Radius of ball (r) = 0.2 m

Angular speed (ω) = 4 rad/s

Wanted : Angular momentum

Solution :

Formula of moment of inertia for uniform sphere :

I = (2/5) m r²

I = moment of inertia (kg m²), m = mass (kg), r = radius (m)

The moment of inertia for uniform sphere :

I = (2/5)(2)(0.2)² = (4/5)(0.04) = 0.032 kg m²

The angular momentum of sphere :

L = I ω = (0.032)(4) = 0.128 kg m²/s

4. A 1-kg particle rotates at a constant angular speed of 2 rad/s. What is the angular speed if the radius of circle is 10 cm.

Known :

Mass of object (m) = 1 kg

The radius of circle (r) = 10 cm = 10/100 = 0.1 m

The angular speed (ω) = 2 rad/s

Wanted : Angular momentum

Solution :

Formula of moment of inertia for particle :

I = m r² = (1)(0.1)² = (1)(0.01) = 0.01 kg m²

Angular momentum :

L = I ω = (0.01)(2) = 0.02 kg m²/s

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1. An object has the moment of inertia of 1 kg m² rotates at a constant angular speed of 2 rad/s. What is the rotational kinetic energy of the object?

Known :

The moment of inertia (I) = 1 kg m²

The angular velocity (ω) = 2 rad/s

Wanted: The rotational kinetic energy (KE)

Solution :

The formula of the rotational kinetic energy :

KE = 1/2 I ω²

KE = the rotational kinetic energy (kg m²/s²), I = the moment of inertia (kg m²), ω = the angular velocity (rad/s)

The rotational kinetic energy :

KE = 1/2 I ω² = 1/2 (1)(2)² = 1/2 (1)(4) = 2 Joule

2. A 20-kg cylinder pulley with a radius of 0.2 m rotates at a constant angular speed of 4 rad/s. What is the rotational kinetic energy of the pulley?

Known :

Mass of cylinder pulley (m) = 20 kg

The radius of cylinder (r) = 0.2 m

The angular speed (ω) = 4 rad/s

Wanted : What is the rotational kinetic energy

Solution ;

Formula of the moment inertia of cylinder :

I = 1/2 m r²

I = the moment of inertia (kg m²), m = mass (kg), r = radius (meter)

The moment of inertia of cylinder pulley :

I = 1/2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

The rotational kinetic energy of the pulley :

KE = 1/2 I ω² = 1/2 (0.4)(4)² = (0.2)(16) = 3.2 Joule

3. A-10 kg ball with radius of 0.1 m rotates at a constant of 10 rad/s. What is the kinetic energy of the ball.

Known :

Mass of ball (m) = 10 kg

Radius of ball (r) = 0.1 m

Angular velocity (ω) = 10 rad/s

Wanted : The rotational kinetic energy

Solution :

Formula of the moment of inertia :

I = (2/5) m r²

I = moment of inertia (kg m²), m = mass (kg), r = radius (m)

Moment of inertia of the ball :

I = (2/5)(10)(0.1)² = (4)(0.01) = 0.04 kg m²

The rotational kinetic energy of the ball :

KE = 1/2 I ω² = 1/2 (0.04)(10)² = (0.02)(100) = 2 Joule

4. A 0.5-kg particle rotates at a constant angular speed of 2 rad/s. What is the rotational kinetic energy of the particle if the radius of circle is 10 cm.

Known :

Mass of particle (m) = 0.5 kg

The radius of ball (r) = 10 cm = 10/100 = 0.1 m

The angular speed (ω) = 2 rad/s

Wanted : The rotational kinetic energy

Solution :

Moment of inertia for particle :

I = m r² = (0.5)(0.1)² = (0.5)(0.01) = 0.005 kg m²

The rotational kinetic energy :

KE = 1/2 I ω² = 1/2 (0.005)(2)² = 1/2 (0.005)(4) = (0.005)(2) = 0.01 Joule

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1. A force F applied to a cord wrapped around a cylinder pulley. The torque is 2 N m and the moment of inertia is 1 kg m², what is the angular acceleration of the cylinder.

Known :

Torque (τ) = 2 N m

The moment of inertia (I) = 1 kg m²

Wanted: The angular acceleration of the cylinder

Solution :

Στ = I α

Στ = net torque, I = moment of inertia, α = angular acceleration

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s²

2. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of the force is 10 N, the radius of the cylinder is 0.2 m and the moment of inertia is 1 kg m2, What is the angular acceleration of the cylinder?

Known :

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

The moment of inertia (I) = 1 kg m²

Wanted: The angular acceleration of the cylinder.

Solution :

τ = F R

τ = torque, F = force, R = radius of cylinder

Torque :

τ = F R = (10 N)(0.2 m) = 2 N m

Στ = I α

Στ = net torque, I = moment of inertia, α = angular acceleration

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s²

3. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of force is 10 N, the radius of cylinder is 0.2 m and the mass of cylinder is 20 kg m^2,. What is the angular acceleration of the cylinder.

Known :

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

Mass of cylinder (M) = 20 kg

Wanted : Angular acceleration of cylinder

Solution :

τ = F R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1⁄2 M R² = 1⁄2 (20)(0.2)² = 1⁄2 (20)(0.04) = 0.4 kg m²

Angular acceleration of cylinder :

α = Στ / I = 2 / 0.4 = 5 rad/s²

4. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The moment of inertia of pulley is 1 kg m² and the radius of pulley is 0.2 m. What is the angular acceleration of the pulley. Acceleration due to gravity is 10 m/s².

Known :

Moment of inertia of pulley (I) = 1 kg m²

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (1 kg)(10 m/s²) = 10 kg m/s² = 10 N

Radius of pulley (R) = 0.2 m

Wanted : Angular acceleration

Solution :

Torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1 kg m²

Angular acceleration :

α = Στ / I = 2 / 1 = 2 rad/s²

5. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The mass of pulley is 20 kg and the radius of pulley is 0,2 m. What is the angular acceleration of the pulley and the free fall acceleration of the block. Acceleration due to gravity is 10 m/s².

Known :

Mass of pulley (M) = 20 kg

Radius of pulley (R) = 0,2 m

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (1 kg)(10 m/s²) = 10 kg m/s² = 10 N

Wanted : the angular acceleration of the pulley and the free fall acceleration of the block.

Solution :

The torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

The moment of inertia of cylinder pulley :

I = 1⁄2 M R² = 1⁄2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

The angular acceleration of the pulley :

α = Στ / I = 2 / 0.4 = 5 rad/s²

The free fall acceleration of the block :

a = R α = (0.2)(5) = 1 m/s²

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The moment of inertia of the particle

1. A 100-gram ball connected to one end of a cord with a length of 30 cm. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass.

Known :

The axis of rotation at AB

Mass ball (m) = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis rotation (r) = 30 cm = 0.3 m

Wanted: Moment of inertia of ball (I)

Solution :

I = m r² = (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. A 100-gram ball, m₁, and a 200-gram ball, m₂, connected by a rod with a length of 60 cm. The mass of the rod is ignored. The axis of rotation located at the center of the rod. What is the moment of inertia of the balls about the axis of rotation?

Known :

Mass of ball 1 (m₁) = 100 gram = 100/1000 = 0.1 kg

The distance of ball 1 and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

Mass of ball (m₂) = 200 gram = 200/1000 = 0.2 kg

The distance of ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

Wanted : moment of inertia of the balls

Jawab :

I = m₁ r₁² + m₂ r₂²

I = (0.1 kg)(0.3 m)² + (0.2 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.2 kg)(0.09 m²)

I = 0.009 kg m² + 0.018 kg m²

I = 0.027 kg m²

3. A 200-gram ball, m₁ and a 100-gram ball, m_2, connected by a rod with length of 60 cm. Ignore rod’s mass. The axis of rotation located at ball m₂. What is the moment of inertia of the balls. Ignore rod’s mass.

Known :

Mass of ball 1 (m₁) = 200 gram = 200/1000 = 0.2 kg

The distance between ball 1 and the axis of rotation (r₁) = 60 cm = 60/100 = 0.6 m

Mass of ball 2 (m₂) = 100 gram = 100/1000 = 0.1 kg

The distance between ball 2 and the axis of rotation (r₂) = 0 m

Wanted : Moment of inertia of the balls

Solution :

I = m₁ r₁² + m₂ r₂²

I = (0.2 kg)(0,6 m)² + (0.2 kg)(0)²

I = (0.2 kg)(0.36 m²) + 0

I = 0.072 kg m²

4. The mass of each ball is 100 gram, connected by cord. The length of cord is 60 cm and the width of cord is 30 cm. What is the moment of inertia of the balls about the axis of rotation. Ignore cord’s mass.

Known :

Mass of ball = m₁ = m₂ = m₃ = m₄ = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

The distance between ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

The distance between ball 3 and the axis of rotation (r₃) = 30 cm = 30/100 = 0.3 m

The distance between ball 4 and the axis of rotation (r₄) = 30 cm = 30/100 = 0.3 m

Known : Moment of inertia

Solution :

I = m₁ r₁² + m₂ r₂² + m₃ r₃² + m₄ r₄²

I = (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²)

I = 0.036 kg m²

The moment of inertia of rigid object

5. What is the moment of inertia of a 2-kg long uniform rod with length of 2 m. The axis of rotation located at the center of the rod.

Known :

Mass of rod (M) = 2 kg

The length of rod (L) = 2 m

Wanted: Moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at the center of long uniform rod :

I = (1/12) M L²

I = (1/12) (2 kg)(2 m)²

I = (1/12) (2 kg)(4 m²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

6. What is the moment of inertia of a 2-kg long uniform rod with a length of 2 m? The axis of rotation located at one end of the rod.

Known :

Mass of rod (M) = 2 kg

The length of rigid rod (L) = 2 m

Wanted: Moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at one end of the rod :

I = (1/3) M L²

I = (1/3) (2 kg)(2 m)²

I = (1/3) (2 kg)(4 m²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

7. A 10-kg solid cylinder with a radius of 0.1 m. The axis of rotational located at the center of the solid cylinder, shown in the figure below. What is the moment of inertia of the cylinder?

Known :

Mass of solid cylinder (M) = 10 kg

Radius of cylinder (L) = 0.1 m

Wanted: The moment of inertia

Wanted: The moment of inertia

Solution :

The formula of moment inertia when the axis of rotation located at the center of cylinder :

I = (1/2) M R²

I = (1/2) (10 kg)(0.1 m)²

I = (1/2) (10 kg)(0.01 m²)

I = (1/2)(0.1 kg m²)

I = 0.05 kg m²

8. A 20-kg uniform sphere with the length of 0.1 m. The axis of rotation located at the center of the sphere shown in the figure below.

Known :

Mass of sphere (M) = 20 kg

The radius of sphere (L) = 0.1 m

Wanted: a moment of inertia

Solution :

The formula of the moment of inertia when the axis of rotation located at the center of the sphere :

I = (2/5) M R²

I = (2/5)(20 kg)(0.1 m)²

I = (2/5)(20 kg)(0.01 m²)

I = (2/5)(0.2 kg m²)

I = 0.4/5 kg m²

I = 0.08 kg m²

9. A 2-kg rectangular thin plate with a length of 0.5 m and width of 0.2 m. The axis of rotation located at the center of the rectangular plat shown in the figure below. What is the moment of inertia of the rectangular?

Known :

Mass of rectangular plat (M) = 2 kg

The length of plat (a) = 0.5 m

The width of plat (b) = 0.2 m

Wanted : Moment of inertia

Solution :

Formula of moment of inertia when the axis of rotation located at the center of plat :

I = (1/12) M (a² + b²)

I = (1/12)(2)(0.5² + 0.2²)

I = (2/12)(0.25 + 0.04)

I = (1/6)(0.29)

I = 0.29/6 kg m²

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