Physics

Linear expansion is experienced only by solid objects; volume expansion is experienced by all objects, both solid, liquid, and gas. The equation of volume expansion is similar to the equation of linear expansion.

Description: V_o = Initial volume, V = Final volume, ΔV = V – V_o = The change in volume, T_o = Initial temperature, T = Final temperature, ΔT = T – T_o = The change in temperature, β = the coefficient of volume expansion. Unit of β = (C^o) ^-1

The above volume expansion equation applies only when the changes in the volume of the objects (both solid, liquid, and gas) are smaller than the original volume of the object. If the change in volume of an object is greater than the initial volume of the object, the equation of the volume expansion does not give the right results. Usually, the changes in volume experienced by solid objects are not too large. Conversely, the coefficient of volume expansion of the liquid and gas is large. The coefficient of volume expansion for gaseous substances is also easy to change if the temperature changes. Therefore the formula above is used only for the expansion of solid objects.

1. At 30 ^oC the volume of an aluminum sphere is 30 cm³. The coefficient of linear expansion is 24 x 10^{-6 o}C^-1. If the final volume is 30.5 cm³, what is the final temperature of the aluminum sphere?

Known :

The coefficient of linear expansion (α) = 24 x 10^{-6 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 24 x 10^{-6 o}C^-1 = 72 x 10^{-6 o}C^-1

The initial temperature (T₁) = 30^oC

The initial volume (V₁) = 30 cm³

The final volume (V₂) = 30.5 cm³

The change in volume (ΔV) = 30.5 cm³ – 30 cm³ = 0.5 cm³

Wanted : The final temperature (T₂)

Solution :

ΔV = β (V₁)(ΔT)

ΔV = β (V₁)(T₂ – T₁)

0.5 cm³ = (72 x 10^{-6 o}C^-1)(30 cm³)(T₂ – 30^oC)

0.5 = (2160 x 10^-6)(T₂ – 30)

0.5 = (2.160 x 10^-3)(T2 – 30)

0.5 = (2.160 x 10^-3)(T₂ – 30)

0.5 / (2.160 x 10^-3) = T₂ – 30

0.23 x 10³ = T₂ – 30

0.23 x 1000 = T₂ – 30

230 = T₂ – 30

230 + 30 = T₂

T₂ = 260^oC

2. The coefficient of linear expansion of a metal sphere is 9 x 10^{-6 o}C^-1. The internal diameter of the metal sphere at 20 ^oC is 2.2 cm. If the final diameter is 2.8 cm, what is the final temperature!

Known :

The coefficient of linear expansion (α) = 9 x 10^{-6 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10^{-6 o}C^-1 = 27 x 10^{-6 o}C^-1

The initial temperature (T₁) = 20^oC

The initial diameter (D₁) = 2.2 cm

The final diameter (D₂) = 2.8 cm

The initial radius (r₁) = D₁ / 2 = 2.2 cm³ / 2 = 1.1 cm³

The final radius (r₂) = D₂ / 2 = 2.8 cm³ / 2 = 1.4 cm³

The initial volume (V₁) = 4/3 π r₁³ = (4/3)(3.14)(1.1 cm)³ = (4/3)(3.14)(1.331 cm³) = 5.57 cm³

The final volume (V₂) = 4/3 π r₂³ = (4/3)(3.14)(1.4 cm)³ = (4/3)(3.14)(2.744 cm³) = 11.48 cm³

The change in volume (ΔV) = 11.48 cm³ – 5.57 cm³ = 5.91 cm³

Wanted : The final temperature (T₂)

Solution :

ΔV = β (V₁)(ΔT)

5.91 cm³ = (27 x 10^{-6 o}C^-1)(5.57 cm³)(T₂ – 20^oC)

5.91 = (150.39 x 10^-6)(T₂ – 20)

5.91 / 150.39 x 10^-6 = T₂ – 20

0.039 x 10⁶ = T₂ – 20

39 x 10³ = T₂ – 20

39,000 = T₂ – 20

39,000 + 20 = T₂

T₂ = 39,020 ^oC

3. A 2000-cm³ aluminum container, filled with water at 0^oC. And then heated to 90^oC. If the coefficient of linear expansion for aluminum is 24 x 10^-6 (^oC)^-1 and the coefficient of volume expansion for water is 6.3 x 10^-4 (^oC)^-1, determine the volume of spilled water.

Known :

The initial volume of the aluminum container and water (V_o) = 2000 cm³ = 2 x 10³ cm³

The initial temperature of the aluminum container and water (T₁) = 0^oC

The final temperature of the aluminum container and water (T₂) = 90^oC

The coefficient of linear expansion for aluminum (α) = 24 x 10^-6 (^oC)^-1

The coefficient of volume expansion for aluminum (γ) = 3α = 3 (24 x 10^-6 (^oC)^-1 ) = 72 x 10^{-6 o}C^-1

The coefficient of volume expansion for water (γ) = 6.3 x 10^-4 (^oC)^-1

Wanted : The volume of spilled water

Solution :

The equation of the volume expansion :

V = V_o + γ V_o ΔT

V – V_o = γ V_o ΔT

ΔV = γ V_o ΔT

V = final volume, V_o = initial volume, ΔV = the change in volume, γ = the coefficient of volume expansion, ΔT = the change in temperature

Calculate the change in volume of the aluminum container :

ΔV = γ V_o ΔT = (72 x 10^-6)(2 x 10³)(90) = 12960 x 10^-3 = 12.960 cm³

Calculate the change in volume of the water :

ΔV = γ V_o ΔT = (6.3 x 10^-4)(2 x 10³)(90) = 1134 x 10^-1 = 113.4 cm³

The change in volume of the water is greater than the aluminum container so that some water spilled.

Calculate the volume of spilled water :

113.4 cm³ – 12.960 cm³ = 100.44 cm³

[wpdm_package id=’702′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, the heat of fusion, the heat of vaporization
9. Energy conservation for heat transfer

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1. At 20 ^oC, the length of a sheet of steel is 50 cm and the width is 30 cm. If the coefficient of linear expansion for steel is 10^{-5 o}C^-1, determine the change in area and the final area at 60 ^oC.

Known :

The initial temperature (T₁) = 20^oC

The final temperature (T₂) = 60^oC

The change in temperature (ΔT) = 60^oC – 20^oC = 40^oC

The initial area (A₁) = length x width = 50 cm x 30 cm = 1500 cm²

The coefficient of linear expansion for steel (α) = 10^{-5 o}C^-1

The coefficient of area expansion for steel (β) = 2α = 2 x 10^{-5 o}C^-1

Wanted : The change in area (ΔA)

Solution :

The change in area (ΔA) :

ΔA = β A₁ ΔT

ΔA = (2 x 10^{-5 o}C^-1)(1500 cm²)(40^oC)

ΔA = (80 x 10^-5)(1500 cm²)

ΔA = 120,000 x 10^-5 cm²

ΔA = 1.2 x 10⁵ x 10^-5 cm²

ΔA = 1.2 cm²

The final area (A₂) :

A₂ = A₁ + ΔA

A₂ = 1500 cm² + 1.2 cm²

A₂ = 1501.2 cm²

2. At 30 ^oC, the area of a sheet of aluminum is 40 cm² and the coefficient of linear expansion is 24 x 10^-6 /^oC. Determine the final temperature if the final area is 40.2 cm².

Known :

The initial temperature (T₁) = 30^oC

The coefficient of linear expansion (α) = 24 x 10^{-6 o}C^-1

The coefficient of area expansion (β) = 2a = 2 x 24 x 10^{-6 o}C^-1 = 48 x 10^{-6 o}C^-1

The initial area (A₁) = 40 cm²

The final area (A₂) = 40.2 cm²

The change in area (ΔA) = 40.2 cm² – 40 cm² = 0.2 cm²

Wanted : Determine the final temperature (T₂)

Solution :

Formula of the change in area (ΔA) :

ΔA = β A₁ ΔT

The final temperature (T₂) :

ΔA = β A₁ (T₂ – T₁)

0.2 cm² = (48 x 10^{-6 o}C^-1)(40 cm²)(T₂ – 30^oC)

0.2 = (1920 x 10^-6)(T₂ – 30)

0.2 = (1.920 x 10^-3)(T₂ – 30)

0.2 = (2 x 10^-3)(T₂ – 30)

0.2 / (2 x 10^-3) = T₂ – 30

0.1 x 10³ = T₂ – 30

1 x 10² = T₂ – 30

100 = T₂ – 30

100 + 30 = T₂

T₂ = 130

The final temperature = 130^oC

3. The radius of a ring at 20 ^oC is 20 cm. If the final radius at 100 ^oC is 20.5 cm, determine the coefficient of area expansion and the coefficient of linear expansion…

Known :

The initial temperature (T₁) = 30^oC

The final temperature (T₂) = 100^oC

The change in temperature (ΔT) = 100^oC – 30^oC = 70^oC

The initial radius (r₁) = 20 cm

The final radius (r₂) = 20.5 cm

Wanted : The coefficient of area expansion (β)

Solution :

The initial area (A₁) = π r₁² = (3.14)(20 cm)² = (3.14)(400 cm²) = 1256 cm²

The final area (A₂) = π r₂² = (3.14)(20.5 cm)² = (3.14)(420.25 cm²) = 1319.585 cm²

The change in area (ΔA) = 1319.585 cm² – 1256 cm² = 63.585 cm²

Formula of the change in area (ΔA) :

ΔA = β A₁ ΔT

The coefficient of area expansion :

ΔA = β A₁ ΔT

63.585 cm² = b (1256 cm²)(70 ^oC)

63.585 = b (87,920 ^oC)

β = 63.585 / 87,920 ^oC

β = 0.00072 /^oC

β = 7.2 x 10^-4 /^oC

β = 7.2 x 10^{-4 o}C^-1

The coefficient of linear expansion (α) :

β = 2 α

α = β / 2

α = (7.2 x 10^-4) / 2

α = 3.6 x 10^{-4 o}C^-1

[wpdm_package id=’698′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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1. A steel is 40 cm long at 20 ^oC. The coefficient of linear expansion for steel is 12 x 10^-6 (C^o)^-1. The increase in length and the final length when it is at 70 ^oC will be…

Known :

The change in temperature (ΔT) = 70^oC – 20^oC = 50^oC

The original length (L₁) = 40 cm

Coefficient of linear expansion for steel (α) = 12 x 10^-6 (C^o)^-1

Wanted : The change in length (ΔL) and the final length (L₂)

Solution :

a) The change in length (ΔL)

ΔL = α L₁ ΔT

ΔL = (12×10^{-6 o}C^-1)(40 cm)(50^oC)

ΔL = (10^-6)(24 x 10³) cm

ΔL = 24 x 10^-3 cm

ΔL = 24 / 10³ cm

ΔL = 24 / 1000 cm

ΔL = 0.024 cm

b) The final length (L₂)

L₂ = L₁ + ΔL

L₂ = 40 cm + 0.024 cm

L₂ = 40.024 cm

2. An iron rod heated from 30 ^oC to 80 ^oC. The final length of iron is 115 cm and the coefficient of linear expansion is 3×10^{-3 o}C^-1. What is the original length and the change in length of the iron ?

Solution :

The change in temperature (ΔT) = 80 ^oC – 30 ^oC = 50 ^oC

The final length (L₂) = 115 cm

The coefficient of linear expansion (α) = 3×10^{-3 o}C^-1

Wanted : the original length (L₁) and the change in length (ΔL)

Solution :

a) The original length (L₁)

Formula of the change in length for the linear expansion :

ΔL = α L₁ ΔT

Formula of the final length :

L₂ = L₁ + ΔL

L₂ = L₁ + α L₁ ΔT

L₂ = L₁ (1 + α ΔT)

115 cm = L₁ (1 + (3.10^{-3 o}C^-1)(50^oC)

115 cm = L₁ (1 + 150.10^-3)

115 cm = L₁ (1 + 0.15)

115 cm = L₁ (1.15)

3. At 25 ^oC, the length of the glass is 50 cm. After heated, the final length of the glass is 50.9 cm. The coefficient of linear expansion is α = 9 x 10^-6 C^-1. Determine the final temperature of the glass…

Known :

The original length (L₁) = 50 cm

The final length (L₂) = 50.09 cm

The change in length (ΔL) = 50.2 cm – 50 cm = 0.09 cm

The coefficient of linear expansion (α) = 9 x 10^{-6 o}C^-1

The original temperature (T₁) = 25^oC

Wanted : The final temperature (T₂)

Solution :

ΔL = α L₁ ΔT

ΔL = α L₁ (T₂ – T₁)

0.09 cm = (9 x 10^{-6 o}C)(50 cm)(T₂ – 25 ^oC)

0.09 = (45 x 10^-5)(T₂ – 25)

0.09 / (45 x 10^-5) = T₂ – 25

0.002 x 10⁵ = T₂ – 25

2 x 10² = T₂ – 25

200 = T₂ – 25

T₂ = 200 + 25

T₂ = 225^oC

The final temperature is 225 ^oC.

4. The original length of metal is 1 meter and the final length is 1.02 m. The change in temperature is 50 Kelvin. Determine the coefficient of linear expansion!

Known :

The initial length (L₁) = 1 meter

The final length (L₂) = 1.02 meter

The change in length (ΔL) = L₂ – L₁ = 1.02 meter – 1 meter = 0.02 meter

The change in temperature (ΔT) = 50 Kelvin = 50^oC

Wanted : The coefficient of linear expansion

Solution :

ΔL = α L₁ ΔT

0.02 m = α (1 m)(50^oC)

0.02 = α (50^oC)

α = 0.02 / 50^oC

α = 0.0004 ^oC^-1

α = 4 x 10^{-4 o}C^-1

[wpdm_package id=’694′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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9 Converting temperature scales (Celsius scale Fahrenheit scale Kelvin scale)

1. 50 ^oC = ….. ^oF ?

Solution

At standard atmospheric pressure, the freezing point of water is 0 ^oC on the Celsius scale and 32 ^oF on the Fahrenheit scale. At standard atmospheric pressure, the boiling point of water is 100 ^oC on the Celsius scale and 212 ^oF on the Fahrenheit scale.

0 ^oC = 32 ^oF and 100 ^oC = 212 ^oF. A change of 5 C^o = a change of 9 F^o.

For a Celsius scale, the distance between 0 ^oC and 100 ^oC divided into 100 equal intervals. For a Fahrenheit scale, the distance between 0 ^oC and 100 ^oC divided into 180 equal intervals.

T^oF = (180/100) T^oC + 32

T^oF = (9/5) T^oC + 32

T^oF = (9/5) 50 + 32

T^oF = (9) 10 + 32

T^oF = 90 + 32

T^oF = 122

50 ^oC = 122 ^oF

2. 86 ^oF = ….. ^oC ?

Solution

T^oC = (100/180)(T^oF – 32)

T^oC = (5/9)(T^oF – 32)

T^oC = (5/9)(86 – 32)

T^oC = (5/9)(54)

T^oC = (5)(6)

T^oC = 30

86 ^oF = 30 ^oC

3. 50^oC = ….. K ?

Solution

T = T ^oC + 273

T = 50 + 273

T = 323

50 ^oC = 323 K

4. 212^oF = ….. K ?

Solution

T^oC = (100/180)(T^oF – 32)

T^oC = (5/9)(T^oF – 32)

T^oC = (5/9)(212 – 32)

T^oC = (5/9)(180)

T^oC = (5)(20)

T^oC = 100

212 ^oF = 100 ^oC + 273

212 ^oF = 373 K

5. x ^oC = x ^oF

x = ….. ?

Solution

1 : Converting Celsius scale into Fahrenheit scale

2 : Converting Fahrenheit scale into Celsius scale

6. 122°F = ….. Celsius

Solution

The conversion between the two temperature scales can be written :

T_C = 5/9 (T_F – 32)

T_C = Temperature in Celsius, T_F = temperature in Fahrenheit

The temperature in Celsius :

T_C = 5/9 (122 – 32) = T_C = 5/9 (90) = 5 (10)

T_C = 50 ^oC

7. The figure below shows the temperature measurement of a liquid with the Fahrenheit scale thermometer! If the temperature of the liquid is measured using a Celsius scale thermometer, then what is the liquid temperature.

Known :

Fahrenheit scale (T_F) = 95^oF

Wanted : Celsius scale

Solution :

At a pressure of 1 atm, the freezing point of water is 0 °C while the Fahrenheit scale is 32 ^oF. Conversely, the boiling point of water for the Celsius scale is 100 ^oC while the Fahrenheit scale is 212 ^oF.

On the Celsius scale, between 0 °C and 100 ° C there are 100 ° while on the Fahrenheit scale between 32 °F to 212 °F there is 180°.

T_C = 100/180 (T_F – 32)

T_C = 5/9 (T_F – 32)

T_C = 5/9 (95 – 32)

T_C = 5/9 (63)

T_C = 315/9

T_C = 35^oC

8. Based on figure below, determine the temperature P on the Celsius thermometer.

Solution

T_C = 100/180 (T_F – 32)

T_C = 5/9 (T_F – 32)

T_C = 5/9 (104 – 32)

T_C = 5/9 (72)

T_C = 360/9

T_C = 40 ^oC

9. If the temperature of Celsius scale as shown in figure below, determine the temperature of Fahrenheit scale as shown in figure below.

Solution :

T^oF = (180/100) T^oC + 32

T^oF = (9/5) T^oC + 32

T^oF = (9/5) 60 + 32

T^oF = (9) 12 + 32

T^oF = 108 + 32

T^oF = 140

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant!

Solution

Hooke’s law formula :

k = F / x

F = force (Newton)

k = spring constant (Newton/meter)

x = the change in length (meter)

Spring constant :

k = 10 / 0.02 = 20 / 0.04

k = 500 N/m

2. Determine the spring constant.

Solution

Spring constant :

k = F / x

k = 5 / 0.01 = 10 / 0.02 = 15 / 0.03 = 20 / 0.04

k = 500 N/m

3. Spring A has the original length of 60 cm and spring B has the original length of 90 cm. Spring A has constant 100 N/m, spring B has constant 200 N/m. The ratio of the change in length of spring A to the change in length of spring B is…

Known :

Constant of spring A (k_A) = 100 N/m

Constant of spring B (k_B) = 200 N/m

Force on spring A (F_A) = F

Force on spring B (F_B) = F

Wanted: Δl_A : Δl_B

Solution :

Hooke’s law formula :

Δl = F / k

Δl = the change in length, F = force, k = constant

The change in length of spring A :

Δl_A = F_A / k_A = F / 100

The change in length of spring B :

Δl_B = F_B / k_B = F / 200

The ratio of the change in length of spring A to the change in length of spring B :

Δl_A : Δl_B

F / 100 : F / 200

1 / 100 : 1 / 200

1 / 1 : 1 / 2

2 : 1

4. A nylon string with original length 20 cm, is pulled by a force of 10 N. The change in length of the string is 2 cm. Determine the magnitude of force if the change in length is 6 cm.

Known :

Force (F) = 10 N

The change in length (Δl) = 2 cm = 0.02 m

Wanted : the magnitude of force (F) if Δl = 0.06 m.

Solution :

Constant :

k = F / Δl

k = 10 / 0.02 = 500 N/m

The magnitude of force (F) if Δl = 0.06 m :

F = k x

F = (500)(0.06)

F = 30 N

[wpdm_package id=’689′]

1. Hooke’s law
2. Stress, strain, Young’s modulus

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Stress Strain Young’s modulus – Problems and Solutions

1. A nylon string has a diameter of 2 mm, pulled by a force of 100 N. Determine the stress!

Known :

Force (F) = 100 N

Diameter (d) = 2 mm = 0.002 m

Radius (r) = 1 mm = 0.001 m

Wanted : The stress

Solution :

Area :

A = π r²

A = (3.14)(0.001 m)² = 0.00000314 m²

A = 3.14 x 10^-6 m²

The stress :

2. A cord has original length of 100 cm is pulled by a force. The change in length of the cord is 2 mm. Determine the strain!

Known :

Original length (l₀) = 100 cm = 1 m

The change in length (Δl) = 2 mm = 0.002 m

Wanted : The strain

Solution :

The strain :

3. A string 4 mm in diameter has original length 2 m. The string is pulled by a force of 200 N. If the final length of the spring is 2.02 m, determine : (a) stress (b) strain (c) Young’s modulus

Known :

Diameter (d) = 4 mm = 0.004 m

Radius (r) = 2 mm = 0.002 m

Area (A) = π r² = (3.14)(0.002 m)²

Area (A) = 0.00001256 m² = 12.56 x 10^-6 m²

Force (F) = 200 N

Original length of spring (l₀) = 2 m

The change in length (Δl) = 2.02 – 2 = 0.02 m

Wanted : (a) The stress (b) The strain c) Young’s modulus

Solution :

(a) The stress

(b) The Strain

(c) Young’s modulus

4. A string has a diameter of 1 cm and the original length of 2 m. The string is pulled by a force of 200 N. Determine the change in length of the string! Young’s modulus of the string = 5 x 10⁹ N/m²

Known :

Young’s modulus (E) = 5 x 10⁹ N/m²

Original length (l₀) = 2 m

Force (F) = 200 N

Diameter (d) = 1 cm = 0.01 m

Radius (r) = 0.5 cm = 0.005 m = 5 x 10^-3 m

Area (A) = π r² = (3.14)(5 x 10^-3 m)² = (3.14)(25 x 10^-6 m²)

Area (A) = 78.5 x 10^-6 m² = 7.85 x 10^-5 m²

Wanted : The change in length (Δl)

Solution :

Young’s modulus formula :

The change in length :

5. A concrete has a height of 5 meters and has unit area of 3 m³ supports a mass of 30,000 kg. Determine (a) The stress (b) The strain (c) The change in height! Acceleration due to gravity (g) = 10 m/s². Young’s modulus of concrete = 20 x 10⁹ N/m²

Known :

Young’s modulus of concrete = 20 x 10⁹ N/m²

Initial height (l₀) = 5 meters

Unit area (A) = 3 m²

Weight (w) = m g = (30,000)(10) = 300,000 N

Wanted : (a) The stress (b) The strain (c) The change in height!

Solution :

(a) The stress

(b) The Strain

(c) The change in height

1. Hooke’s law
2. Stress, strain, Young’s modulus

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1. A ball, attached to the end of a horizontal cord, is revolved in a circle of radius 20 cm. The ball around 360^o each second. Determine the magnitude of the centripetal acceleration!

Known :

Angular speed (ω) = 360^o/second = 1 revolution/second = 6.28 radians/second

Radius (r) = 20 cm = 0.2 m

Wanted : Centripetal acceleration (a_r)

Solution :

a_r = v² / r —> v = r ω

a_r = (r ω)² / r = r² ω² / r

a_r = r ω²

a_s = centripetal acceleration, v = linear velocity, r = radius, ω = angular velocity

The magnitude of the centripetal acceleration :

a_r = r ω² a_r = (0,2 m)(6.28 rad/s)

a_r = 1.256 m/s²

2. A wheel 30 cm in radius rotate at a rate of 180 rpm. Determine the centripetal acceleration of a point on the edge of wheel!

Known :

Radius (r) = 30 cm = 0.3 m

Angular speed (ω) = 180 revolutions / 60 seconds = 3 revolutions / second = (3)(6.28 radians) / second = 18.84 radians/second

Wanted : centripetal acceleration (a_r) of r = 0.3 m

Solution :

The magnitude of the centripetal acceleration :

a_r = r ω²

a_r = (0.3 m)(18.84 rad/s)

a_r = 5.65 m/s²

3. A race car moving on a circular track of radius 50 meters. If car’s speed is 72 km/h, determine the magnitude of the centripetal acceleration!

Known :

Radius (r) = 50 meters

Speed (v) = 72 km/h = (72)(1000 meters) / 3600 seconds = 20 meters/second

Wanted : the magnitude of the centripetal acceleration (a_r)

Solution :

a_r = v² / r = 20² / 50 = 400 / 50 = 8 m/s²

4. A car has the maximum centripetal acceleration 10 m/s², so the car can turn without skidding out of a curved path. If the car is moving at a constant 108 km/h, what is the radius of unbanked curve ?

Known :

Centripetal acceleration (a_r) = 10 m/s²

Car’s speed (v) = 108 km/h = (108)(1000) / 3600 = 30 meters/second

Wanted : radius (r)

Solution :

r = v² / a_r

r = 30² / 10 = 900 / 10 = 90 meters

[wpdm_package id=’433′]

[wpdm_package id=’439′]

1. Converting angle units sample problems with solutions
2. Angular displacement and linear displacement sample problems and solutions
3. Angular velocity and linear velocity sample problems with solutions
4. Angular acceleration and linear acceleration sample problems with solutions
5. Uniform circular motions sample problems with solutions
6. Centripetal acceleration sample problems with solutions
7. Nonuniform circular motions sample problems with solutions

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1. A wheel 30 cm in radius rotates at constant 5 rad/s². What is the magnitude of the linear acceleration of a point located at (a) 10 cm from the center (b) 20 cm from the center (c) on the edge of the wheel?

Known :

Radius (r) = 30 cm = 0.3 m

Angular acceleration (α) = 5 rad/s²

Wanted : linear acceleration (a) r = 0.1 m (b) r = 0.2 m (c) r = 0.3 m

Solution :

Relation between linear acceleration (a) and angular acceleration :

a = r α

(a) linear acceleration, r = 0.1 m

a = (0.1 m)(5 rad/s²) = 0.5 m/s²

(b) linear acceleration, r = 0.2 m

a = (0.2 m)(5 rad/s²) = 1 m/s²

(c) linear acceleration, r = 0.3 m

a = (0.3 m)(5 rad/s²) = 1.5 m/s²

2. A pulley 50 cm in radius. If the linear acceleration of a point located on the edge of the pulley is 2 m/s², determine the angular acceleration of the pulley!

Known :

Radius (r) = 50 cm = 0,5 m

linear acceleration (a) = 2 m/s²

Wanted : the angular acceleration

Solution :

α = a / r = 2 / 0.5 = 4 rad/s²

3. The blades in a blender 20 cm in radius, initially at rest. After 2 seconds, the blades rotates 10 rad/s. Determine the magnitude of linear acceleration (a) a point located at 10 cm from the center (b) a point located at the edge of the blades.

Known :

Radius (r) = 20 cm = 0.2 m

The initial angular velocity (ω_o) = 0

The final angular velocity (ω_t) = 10 radians/second

Time interval (t) = 2 seconds

Wanted : the linear acceleration of a point located at (a) r = 0.1 m (b) r = 0.2 m

Solution :

ω_t = ω_o + α t

10 = 0 + α (2)

10 = 2 α

α = 10 / 2

 α = 5 rad/s

(a) linear acceleration of r = 0.1 m

a = r α = (0.1 m)(5 rad/s²) = 0.5 m/s²

(b) linear acceleration of r = 0.2 m

a = r α = (0.2 m)(5 rad/s²) = 1 m/s²

4. A wheel 20 cm in radius is accelerated for 2 seconds from 20 rad/s to rest. Determine the magnitude of linear acceleration (a) a point located at 10 cm from the center (b) a point located at 10 cm from the center.

Known :

Radius (r) = 20 cm = 0.2 m

The initial angular speed (ω_o) = 20 rad/s

The final angular speed (ω_t) = 0

Time interval (t) = 2 seconds

Wanted : The linear acceleration (a) r = 0.1 m (b) r = 0.2 m

Solution :

ω_t = ω_o + α t

0 = 20 + α (2)

-20 = 2 α

α = -20 / 2

 α = -10 rad/s

Negative sign mean the angular speed is decrease.

(a) linear acceleration of r = 0.1 m

 a = r α = (0.1 m)(-10 rad/s²) = -1 m/s²

(b) linear acceleration of r = 0.2 m

a = r α = (0.2 m)(-10 rad/s²) = -2 m/s²

[wpdm_package id=’429′]

[wpdm_package id=’439′]

1. Converting angle units sample problems with solutions
2. Angular displacement and linear displacement sample problems and solutions
3. Angular velocity and linear velocity sample problems with solutions
4. Angular acceleration and linear acceleration sample problems with solutions
5. Uniform circular motions sample problems with solutions
6. Centripetal acceleration sample problems with solutions
7. Nonuniform circular motions sample problems with solutions

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1. A ball at the end of a string is revolving uniformly in a horizontal circle of radius 2 meters at constant angular speed 10 rad/s. Determine the magnitude of the linear velocity of a point located :

(a) 0.5 meters from the center

(b) 1 meter from the center

(c) 2 meters from the center

Known :

Radius (r) = 0.5 meters, 1 meter, 3 meters

The angular speed = 10 radians/second

Wanted : The linear velocity

Solution :

v = r ω

v = the linear velocity, r = radius, ω = the angular velocity

(a) The linear velocity (v) of a point located at r = 0.5 meters

v = r ω = (0.5 meters)(10 rad/s) = 5 meters/second

(b) The linear velocity (v) of a point located at r = 1 meter

v = r ω = (1 meter)(10 rad/s) = 10 meters/second

(c) The linear velocity (v) of a point located at r = 2 meters

v = r ω = (2 meters)(10 rad/s) = 20 meters/second

2. The blades in a blender rotate at a rate of 5000 rpm. Determine the magnitude of the linear velocity :

(a) a point located 5 cm from the center

(b) a point located 10 cm from the center

Known :

Radius (r) = 5 cm and 10 cm

The angular speed (ω) = 5000 revolutions / 60 seconds = 83.3 revolutions / second = (83.3)(6.28 radian) / second = 523.3 radians / second

Wanted : The magnitude of the linear velocity

Solution :

(a) The magnitude of the linear velocity of a point located 0.05 m from the center

v = r ω = (0.05 m)(523.3 rad/s) = 26 m/s

(b) The magnitude of the linear velocity of a point located 0,1 m from the center

v = r ω = (0.1 m)(523.3 rad/s) = 52 m/s

3. A point on the edge of a wheel 30 cm in radius, around a circle at constant speed 10 meters/second.

What is the magnitude of the angular velocity?

Known :

Radius (r) = 30 cm = 0.3 meters

The linear velocity (v) = 10 meters/second

Wanted : the angular velocity

Solution :

ω = v / r = 10 / 0.3 = 33 radians/second

4. A car with tires 50 cm in diameter travels 10 meters in 1 second. What is the angular speed ?

Known :

Radius (r) = 0.25 meter

The linear speed of a point on the edge of tires (v) = 10 meters/second

Wanted: The angular speed

Solution :

ω = v / r = 10 / 0.25 = 40 radians/second

5. The angular speed of wheel 20 cm in radians is 120 rpm. What is the distance if the car travels in 10 seconds.

Known :

Radius (r) = 20 cm = 0.2 meters

The angular speed = 120 rev / 60 seconds = 2 rev / second = (2)(6.28) radians / second = 12.56 radians / second

Wanted : distance

Solution :

Velocity of the edge of wheel :

v = r ω = (0.2 meters)(12.56 radians/second) = 2.5 meters/second

2.5 meters / second means a point on the edge of wheel travels 2.5 meters each 1 second. After 10 seconds, the point travels 25 meters.

So the distance is 25 meters.

[wpdm_package id=’427′]

[wpdm_package id=’439′]

1. Converting angle units sample problems with solutions
2. Angular displacement and linear displacement sample problems and solutions
3. Angular velocity and linear velocity sample problems with solutions
4. Angular acceleration and linear acceleration sample problems with solutions
5. Uniform circular motions sample problems with solutions
6. Centripetal acceleration sample problems with solutions
7. Nonuniform circular motions sample problems with solutions

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Converting angle units (degree, radian, revolution)

1. ¼ rev = ….. ^o (degree) ?

Solution

1 rev = 360^o

½ rev = 180^o

¼ rev = 90^o

2. ½ rev = …….. rad ?

Solution

1 rev = 2π rad = 2(3.14) rad = 6.28 rad

½ rev = pi rad = 3.14 rad

3. 180^o = ….. rev ?

Solution

360^o = 1 rev

180^o = ½ rev

4. 90^o = ….. rad ?

Solution

360^o = 2π rad = 2(3.14) rad = 6.28 rad

180^o = π rad = 3.14 rad

90^o = ½ π rad = ½ (3.14) = 1.57

5. 60 rad = ….. rev ?

Solution

6.28 rad = 1 rev

60 rad/6.28 = 9.55 rev

6. 40 rad= ….. ^o ?

Solution

6.28 rad = 360^o

40 rad/6.28 = (6.37)(360^o) = 2292.99^o

Angular displacement and linear displacement

1. A bike wheel 60 cm in diameter rotates 10 radians. What is the linear displacement of a point on the edge of the wheel?

Known :

Radius (r) = 30 cm = 0.3 m

Angle (θ) = 10 radians

Wanted : linear displacement (l)

Solution :

l = r θ

l = (0.3 m)(10 rad)

l = 3 meters

2. A wheel 50 cm in radius rotates 360^o. What is the linear displacement of a point on the edge of the wheel ?

Known :

Radius (r) = 50 cm = 0.5 meters

Angle (θ) = 360^o = 6.28 radians

Wanted : linear displacement (l)

Solution :

l = r θ

l = (0.5 m)(6.28 rad)

l = 3.14 meters

3. A wheel 50 cm in radius rotates 2 revolutions. What is the linear displacement of a point on the edge of the wheel ?

Known :

Radius (r) = 50 cm = 0,5 m

Angle (θ) = 2 revolutions = (2)(6.28 radians) = 12.56 radians

Wanted : linear displacement (l) ?

Solution :

l = r θ

l = (0.5 m)(12.56 rad)

l = 6.28 m

4. A point on the edge of a wheel 2 meters in radius, moves 100 meters. Determine the angular displacement.

Known :

Radius (r) = ½ (diameter) = ½ (2 meters) = 1 meter

linear displacement (l) = 100 meters

Solution :

(a) Angular displacement (in radian)

θ = s / r = 100 / 1 = 100 radians

(b) Angular displacement (in degrees)

1 radian = 360^o

100 radians = 100(360^o) = 36,000 radians

(c) Angular displacement (in revolution)

6.28 radians = 1 revolution

36,000 / 6.28 = 5732,484 revolutions

5. A particle round a circle 10 meters and rotates 180^o. What is the radius ?

Known :

Linear displacement (l) = 10 meters

Angle (θ) = 180^o = 3.14 radians

Wanted : radius (r)

Solution :

r = l / θ = 10 / 3.14 = 3.18 meters

1. Converting angle units sample problems with solutions
2. Angular displacement and linear displacement sample problems and solutions
3. Angular velocity and linear velocity sample problems with solutions
4. Angular acceleration and linear acceleration sample problems with solutions
5. Uniform circular motions sample problems with solutions
6. Centripetal acceleration sample problems with solutions
7. Nonuniform circular motions sample problems with solutions

Read more