Solved Problems in Basic Physics

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

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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

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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

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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.

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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

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1. A wheel 1 meter in radius accelerates uniformly at 2 rad/s². Determine the angular acceleration and the angular speed of the wheel, 2 seconds later.

Known :

Radius (r) = 1 meter

Angular acceleration (α) = 2 rad/s²

Wanted: angular acceleration and angular speed after 2 seconds.

Solution :

(a) Angular acceleration in 2 seconds

Angular acceleration is constant, thus after 2 seconds, angular acceleration of the wheel is 2 rad/s².

(b) Angular speed in 2 seconds

Angular acceleration 2 rad/s² means the angular speed increases 2 radians/second each 1 second. After 1 second, angular speed = 2 radians/second. After 2 seconds, angular speed = 4 radians/second.

2. A particle accelerates uniformly from rest to 60 rpm in 10 seconds. Determine the magnitude of angular acceleration!

Known :

The initial angular velocity (ω_o) = 0

The final angular velocity (ω_t) = 60 rpm = 60 revolutions / 60 seconds = 1 revolution / second = 6,28 radians/second

Time interval (t) = 10 seconds

Wanted : Angular acceleration (α)

Solution :

ω_o = the initial angular velocity, ω_t = the final angular velocity, α = the angular acceleration, t = time interval, θ = angle.

ω_t = ω_o + α t

6.28 = 0 + α (10)

6.28 = 10 α

α = 6.28 / 10

α = 0.628 rad/s²

The magnitude of the angular acceleration = 0.628 rad/s²

3. An object slows down from 20 rad/s to 10 rad/s in 4 seconds. Determine the magnitude of angular acceleration!

Known :

Time interval (t) = 4 seconds

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

The final angular velocity (ω_t) = 10 rad/s

Wanted : the magnitude of the angular acceleration (α)

Solution :

ω_t = ω_o + α t

10 = 20 + α (4)

10 – 20 = 4 α

-10 = 4 α

α = -10 / 4

α = – 2.5 rad/s²

The magnitude of the angular acceleration is -2.5 rad/s². Negative sign means the object is decelerating. Acceleration = the angular speed increases, deceleration = the angular speed decreases.

4. An object is accelerated for 2 seconds from 10 rad/s to 2 rad/s². Determine angle rounded by the object!

Known :

the initial angular velocity (ω_o ) = 10 rad/s

the angular acceleration (α) = 2 rad/s²

time interval (t) = 2 seconds

Wanted : angle (θ)

Solution :

θ = ω_o + ½ α t²

θ = (10)(2) + ½ (2)(2²)

θ = 20 + (1)(4) = 20 + 4

θ = 24 radians

5. A car’s wheel slows down from 20 rad/s to rest after round 20 radians. Determine the magnitude of the angular acceleration of the wheel!

Known :

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

the final angular speed (ω_t) = 0

Angle (θ) = 20 radians

Wanted : the magnitude of the angular acceleration (α)

Solution :

ω_t² = ω_o² + 2 α θ

0 = 20² + 2 α (20)

0 = 400 + 40 α

400 = – 40 α

α = – 400 / 40

α = – 10 rad/s²

6. A rod PQ with length of 60 cm rotated about point Q as the axis of rotation and PQ as the radius of circle. The rod PQ accelerated from rest to 0.3 rad/s². What is the linear speed of point P at t = 10 seconds, if the angular initial position is 0.

Known :

Length of rod PQ = radius of circle (r) = 60 cm = 60/100 m = 0.60 m

The initial angular speed (ω_o) = 0 rad/s

Angular acceleration (α) = 0.3 rad s^-2

The initial angular position (θ_o) = 0

Wanted : Linear speed (v) of point P at t = 10 seconds

Solution :

The final angular speed after 10 seconds :

ω_t = ω_o + α t = 0 rad/s + (0.3 rad s^-2)(10 s) = 3 rad/s

The final linear speed after 10 seconds :

v = r ω = (0.6 m)(3 rad/s) = 1.8 m/s

7. An object rotates with the initial speed of 4 rad/s and the angular acceleration is 0.5 rad/s². What is the speed of object after 4 seconds.

Known :

The initial angular speed (ω_o) = 4 rad/s

Angular acceleration (α) = 0.5 rad/s²

Time interval (t) = 4 seconds

Wanted : Object’s speed after 4 seconds (ω_t)

Solution :

ω_t = ω_o + α t

ω_t = 4 + (0.5)(4)

ω_t = 4 + 2

ω_t = 6 rad/s

8. A wall clock with diameter of 10 cm has three needles, each to show the hours, minutes and seconds. Comparison of the number of rounds of the hour needle: the minute needle: the second needle.

A. 1 : 3 : 180

B. 1 : 12 : 720

C. 4 : 12 : 180

D. 4 : 12 : 720

Known :

1 hour = 60 minutes

12 hours = (12)(60 minutes) = 720 minutes

Angular speed of the hour needle = 1 revolution / 12 hours = 1 revolution / 720 minutes

Angular speed of the minutes needle = 1 revolution / 1 hour = 1 revolution / 60 minutes

Angular speed of second needle = 1 revolution / 1 minute

Wanted: Comparison of the number of rounds of the hour needle: the minute needle: the second needle

Solution :

The equation of circular motion :

Angular speed = number of revolution / time interval

Number of revolution = angular speed x time interval

In the same time interval, for example, 1 minute, how many revolution of hour needle, minute needle, and the second needle.

Number of revolution of the hour needle = angular speed x time interval = (1 revolution / 720 minutes)(1 minute) = 1/720 revolutions

Number of revolution of the minute needle = angular speed x time interval = (1 revolution / 60 minutes)(1 minute) = 1/60 revolutions

Number of revolution of the second needle = angular speed x time interval = (1 revolution / 1 minute)(1 minute) = 1/1 revolution

Comparison of a number of revolutions :

Number of revolution of the hour needle: number of revolution of minute needle : number of revolution of the second needle.

1/720 : 1/60 : 1/1

1/720 : 12/720 : 720/720

1 : 12 : 720

The correct answer is B.

9. A ball tied with a rope. The ball is rotated so that it moves in a circular plane parallel to the surface of the earth. In this motion, the ball accelerates because…..

A. Friction of air

B. Weight of ball

C. Tension force

D. Force of gravity

Solution :

Newton’s second law of motion states that an object is accelerated if there is a resultant force. The ball is connected to the rope and when the rope rotated, the ball also rotates. When the ball rotates (the ball moves in a circle), the ball undergoes centripetal acceleration. All the moving objects are circular centripetal acceleration. Centripetal acceleration is caused by centripetal force. The centripetal force for this case is the tension force.

The correct answer is C.

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[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. An object moves in a circle with the constant angular speed of 10 rad/s. Determine (a) Angular speed after 10 seconds (b) Angular displacement after 10 seconds.

Known :

Angular speed (ω) =10 rad/s

Wanted :

(a) Angular speed (ω) after 10 seconds.

(b) Angle (θ) after 10 seconds

Solution :

(a) Angular speed (ω) after 10 seconds

Object in uniform circular motion so that angular speed is constant, 10 rad/s.

(b) Angular displacement (θ)

Constant angular speed 10 radians/second means the object around 10 radians each second. After 10 seconds, the object around 10 x 10 radians = 100 radians.

2. A particle moves in a circle with the constant speed of 10 m/s. Radius of circle = 1 meter. Determine (a) Particle’s speed after 5 seconds (b) Particle’s displacement after 5 seconds (c) Centripetal acceleration.

Known :

Radius of circle (r) = 1 meter

Particle’s speed (v) = 10 m/s

Solution :

(a) Particle’s speed after 5 seconds

The motion of object is in the uniform circular motion so that speed is constant, 10 m/s.

(b) Particle’s displacement after 5 seconds

10 meters/second means each 1 second, particle’s displacement = 10 meters. After 5 seconds, particle’s displacement = 5 x 10 meters = 50 meters.

(c) Centripetal acceleration (a_r)

a_r = v² / r = 10² / 1 = 100 / 1 = 100 m/s²

3. A ball attached to one end of a cord, is revolved in a circle with radius of 2 meters at the constant speed of 60 rpm. Determine (a) the magnitude of the angular speed after 2 seconds (b) the angular displacement after 1 minute.

Known :

Radius of circle (r) = 2 meters

Angular speed (ω) = 60 rpm = 60 revolutions / 1 minute

= 60 revolutions / 60 seconds = 1 revolution / second = 2π radians / second

= 2(3.14) radians / second= 6.28 radians / second

Solution :

(a) Angular speed (ω) after 2 seconds

The angular speed is constant so after 2 seconds, the angular speed (ω) = 6.28 radians / second

(b) Angular displacement (θ)

The angular speed = 1 revolution/second means each 1 second, ball experience 1 revolution. After 60 seconds, ball moves 60 revolutions.

The angular speed = 6.28 radians/second means each 1 second, the ball moves with the angle of 6.28 radians. After 60 seconds, the ball moves 376.8 radians.

4. A bike wheel rotates 120 revolutions in 60 seconds. What is the angular speed?

Solution :

(a) revolutions per minute (rpm)

120 revolutions / 60 seconds = 120 revolutions / 1 minute = 120 revolutions / minute = 120 rpm

(b) degrees per second (^o/s)

1 revolution = 360^o, 120 revolutions = 43200^o

120 revolutions / 60 seconds = (120)(360^o) / 60 seconds = 43200^o / 60 seconds = 720^o/second

(c) radians per second (rad/s)

1 revolution = 6.28 radians

120 revolutions / 60 seconds = (120)(6.28) radians / 60 seconds = 753.6 radians / 60 seconds = 12.56 radians/second.

[wpdm_package id=’432′]

[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 0.1-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 50 cm and ball’s angular speed is 4 rad s^-1. What is the magnitude of the centripetal force?

Known :

Mass (m) = 100 gram = 100/1000 kg = 1/10 kg = 0.1 kg

Angular speed (ω) = 4 radians/second

Radius (r) = 50 cm = 50/100 m = 0.5 m

Wanted : Centripetal force

Solution :

Centripetal force is net force which produces centripetal acceleration :

∑F = m a_r

∑F = m v²/r = m ω² r

∑F = net force = centripetal force, m = mass, v = speed, ω = angular speed, r = radius

∑F = m ω² r = (0.1)(4)² (0.5) = (0.1)(16)(0,5) = 0.8 Newton

2. A ball is revolving uniformly in a horizontal circle. If the speed changed to four times the initial value, what is the magnitude of centripetal force…..

Known :

Mass = m

Speed = v

Initial speed = v_o

Radius (r) = r

Wanted: Magnitude of centripetal force

Solution :

3. A banked curve of radius R is designed so that a car traveling at speed 12 ms^–1 can negotiate the turn safely. The coefficient of static friction between car and road = 0.4. What is radius R. Acceleration due to gravity (g) = 10 ms^–2.

Known :

Speed (v) = 12 m/s

Coefficient of static friction (μ_s) = 0.4

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

Wanted: Radius (R)

Solution :

[wpdm_package id=’501′]

1. Mass and weight
2. Normal force
3. Newton’s second law of motion
4. Friction force
5. Motion on a horizontal surface without friction force
6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
7. Motion on the inclined plane without friction force
8. Motion on the rough inclined plane with the friction force
9. Motion in an elevator
10. The motion of bodies connected by cord and pulley
11. Two bodies with the same magnitude of accelerations
12. Rounding a flat curve – dynamics of circular motion
13. Rounding a banked curve – dynamics of circular motion
14. Uniform motion in a horizontal circle
15. Centripetal force in uniform circular motion

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