How to Calculate and Solve for Linear Momentum | Motion

The image above represents linear momentum.

To compute for linear momentum, two essential parameters are needed and these parameters are mass (m) and velocity (v).

The formula for calculating linear momentum:

p = mv

Where;

p = Momentum
m = Mass
v = Velocity

Let’s solve an example;
Find the momentum when the mass is 8 and the velocity is 12.

This implies that;

m = Mass = 8
v = Velocity = 12

p = mv
p = 8 x 12
p = 96

Therefore, the momentum is 96 Kgm/s.

Calculating the Mass when the Momentum and the Velocity is Given.

m = p / v

Where;

m = Mass
p = Momentum
v = Velocity

Let’s solve an example;
Find the mass when the momentum is 40 and the velocity is 18.

This implies that;

p = Momentum = 40
v = Velocity = 18

m = p / v
m = 40 / 18
m = 2.22

Therefore, the mass is 2.22 kg.

Continue reading How to Calculate and Solve for Linear Momentum | Motion

How to Calculate and Solve for Velocity | Motion

The image above represents velocity.

To compute for velocity, three essential parameters are needed and these parameters are angular velocity (ω), amplitude (A) and displacement (y).

The formula for calculating velocity:

v = ω√(A² – y²)

Where;

v = Velocity
ω = Angular Velocity
A = Amplitude
y = Displacement

Let’s solve an example;
Find the velocity when the angular velocity is 7, amplitude is 10 and displacement is 13.

This implies that;

ω = Angular Velocity = 7
A = Amplitude = 10
y = Displacement = 13

v = ω√(A² – y²)
v = 7 x √(10² – 13²)
v = 7 x √(100 – 169)
v = 7 x √(-69)

i signifies a complex number = √(-1).

v = 7 x √(69) x √(-1)
v = 7 x √(69)i
v = 7 x 8.30i
v = 7(8.30i)

Therefore, the velocity is 7(8.30i) m/s.

Calculating the Angular Velocity when the Velocity, the Amplitude and the Displacement is Given.

ω = v / √(A2 – y2)

Where;

ω = Angular Velocity
v = Velocity
A = Amplitude
y = Displacement

Let’s solve an example;
Find the angular velocity when the velocity is 28, the amplitude is 14 and the displacement is 10.

This implies that;

v = Velocity = 28
A = Amplitude = 14
y = Displacement = 10

ω = v / √(A2 – y2)
ω = 28 / √(142 – 102)
ω = 28 / √(196 – 100)
ω = 28 / √(96)
ω = 28 / 9.79
ω = 2.86

Therefore, the angular velocity is 2.86.

Continue reading How to Calculate and Solve for Velocity | Motion

How to Calculate and Solve for Velocity | Motion

The image above represents velocity.

To compute for velocity, two essential parameters are needed and these parameters are angular velocity (ω) and radius (r).

The formula for calculating velocity:

v = ωr

Where;

v = Velocity
ω = Angular Velocity
r = Radius

Let’s solve an example;
Find the velocity when the angular velocity is 17 with a radius of 6.

This implies that;

ω = Angular Velocity = 17
r = Radius = 6

v = ωr
v = 17 x 6
v = 102

Therefore, the velocity is 102 m/s.

Calculating the Angular Velocity when the Velocity and the Radius is Given.

ω =v / r

Where;

ω = Angular Velocity
v = Velocity
r = Radius

Let’s solve an example;
Given that the velocity is 30 with a radius of 3. Find the angular velocity?

This implies that;

v = Velocity = 30
r = Radius = 3

ω = v / r
ω = 30 / 3
ω = 10

Therefore, the angular velocity is 10.

Continue reading How to Calculate and Solve for Velocity | Motion

How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image above represents the maximum velocity to avoid overturning of a vehicle moving along a level circular path.

To compute for the maximum velocity, four essential parameters are needed and these parameters are Acceleration due to Gravity (g), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the maximum velocity:

vmax = √(gra / h)

Where:
vmax = Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path
g = Acceleration due to Gravity
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the maximum velocity when the Acceleration due to Gravity (g) is 10.2, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 14, Radius of Circular Path (r) is 22 and Half of the Distance between the Centre Lines of the Wheel (a) is 32.

This implies that;
g = Acceleration due to Gravity = 10.2
h = Height of Centre of Gravity of the Vehicle from Ground Level = 14
r = Radius of Circular Path = 22
a = Half of the Distance between the Centre Lines of the Wheel = 32

vmax = √(gra / h)
vmax = √((10.2)(22)(32)/14)
vmax = √((7180.79)/14)
vmax = √(512.91)
vmax = 22.647

Therefore, the maximum velocity to avoid Overturning of a Vehicle moving along a Level Circular Path is 22.647 m/s.

Continue reading How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image represents reaction at the inner wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the inner wheel of a vehicle moving along a level circular path:

RA = mg / 2[1 – v²h / gra]

Where:
RA = Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 13, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 11, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 5, Radius of Circular Path (r) is 7 and Half of the Distance between the Centre Lines of the Wheel (a) is 3.

This implies that;
m = Mass of the Vechicle = 13
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 11
h = Height of Centre of Gravity of the Vehicle from Ground Level = 5
r = Radius of Circular Path = 7
a = Half of the Distance between the Centre Lines of the Wheel = 3

RA = mg / 2[1 – v²h / gra]
RA = 13(9.8) / 2[1 – (11)²(5) / (9.8)(7)(3)]
RA = 127.4 / 2[1 – (121)(5) / 205.8]
RA = 63.7[1 – 605 / 205.8]
RA = 63.7[1 – 2.939]
RA = 63.7[-1.939]
RA = -123.56

Therefore, the reaction at the inner wheel of a vehicle moving along a level of circular path is -123.56 N.

Continue reading How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for Road Bank Angle, Velocity and Radius of a Body in Motion of Circular Path | The Calculator Encyclopedia

The image represents road bank angle in circular motion.

To compute for the road bank angle, three essential parameters are needed and these parameters are velocity (v), acceleration due to gravity (g) and radius (r).

The formula for calculating the road bank angle;

θ = tan-1( / gr)

Where;
θ = Road Bank Angle
v = Velocity
g = Acceleration due to Gravity
r = Radius

Let’s solve an example;
Find the road bank angle where the acceleration due to gravity is 9.8, velocity is 35 and radius is 18.

This implies that;
v = Velocity = 35
g = Acceleration due to Gravity = 9.8
r = Radius = 18

θ = tan-1( / gr)
θ = tan-1(35² / (9.8)(18))
θ = tan-1(1225 / 176.4)
θ = tan-1(6.94)
θ = 81.81°

Therefore, the road bank angle is 81.81°.

Calculating the Velocity when Road Bank Angle, Acceleration due to Gravity and Radius is Given.

v = √gr.tan θ

Where;
v = Velocity
θ = Road Bank Angle
g = Acceleration due to Gravity
r = Radius

Let’s solve an example;
Given that the road bank angle is 50, radius is 15 and acceleration due to gravity is 9.8. Find the velocity?

This implies that;
θ = Road Bank Angle = 50
g = Acceleration due to Gravity = 9.8
r = Radius = 15

v = √gr.tan θ
v = √(9.8 x 15)(tan 50)
v = √(147)(1.1917)
v = √175.1799
v = 13.235

Therefore, the velocity is 13.235.

Continue reading How to Calculate and Solve for Road Bank Angle, Velocity and Radius of a Body in Motion of Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

The image represents reaction at the outer wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the outer wheel of a vehicle moving along a level circular path:

RB = mg / 2[1 + v²h/gra]

Where;
RB = Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 12, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 28, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 16, Radius of Circular Path (r) is 8 and Half of the Distance between the Centre Lines of the Wheel (a) is 4.

This implies that;
m = Mass of the Vechicle = 12
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 28
h = Height of Centre of Gravity of the Vehicle from Ground Level = 16
r = Radius of Circular Path = 8
a = Half of the Distance between the Centre Lines of the Wheel = 4

RB = mg / 2[1 + v²h/gra]
RB = 12 x 9.8 / 2[1 + 28² x 16/9.8 x 8 x 4]
RB = 117.60 / 2[1 + 784 x 16/313.6]
RB = 58.80[1 + 40]
RB = 58.80[41]
RB = 58.80[41]
RB = 2410.8

Therefore, the reaction of the outer wheel of a vehicle moving along a level circular path is 2410.8 N.

Continue reading How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

How to Calculate and Solve for Maximum Velocity, Coefficient of Friction and Radius of a Body in Motion of Circular Path | Nickzom Calculator

The image above represents maximum velocity in circular motion.

To compute for the maximum velocity, three essential parameters are needed and these parameters are coefficient of friction (μ), radius (r) and acceleration due to gravity (g).

The formula for calculating maximum velocity:

Vmax = √(μgr)

Where;
Vmax = maximum velocity
μ = coefficient of friction
r = radius
g = acceleration due to gravity

Let’s solve an example;
Find the maximum velocity when the coefficient of friction is 14 with a radius of 7 and acceleration due to gravity of 9.8.

This implies that;
μ = coefficient of friction = 14
r = radius = 7
g = acceleration due to gravity = 9.8

Vmax = √(μgr)
Vmax = √(14 x 7 x 9.8)
Vmax = √(960.40)
Vmax = 30.99

Therefore, the maximum velocity is 30.99 m/s.

Calculating the Coefficient of Friction when the Maximum Velocity, Radius and Acceleration due to Gravity is Given.

μ = Vmax2 / gr

Where;
μ = coefficient of friction
Vmax = maximum velocity
r = radius
g = acceleration due to gravity

Let’s solve an example;
Find the coefficient of friction with a maximum velocity of 120, radius of 15 and acceleration due to gravity is 9.8?

This implies that;
Vmax = maximum velocity = 120
r = radius = 15
g = acceleration due to gravity = 9.8

μ = Vmax2 / gr
μ = 1202 / 15 x 9.8
μ = 14400 / 147
μ = 97.96

Therefore, the coefficient of friction is 97.96.

Continue reading How to Calculate and Solve for Maximum Velocity, Coefficient of Friction and Radius of a Body in Motion of Circular Path | Nickzom Calculator

How to Calculate and Solve for Superelevation, Guage of Track, Velocity and Radius of a Body in Circular Path Motion | The Calculator Encyclopedia

The image above represents a body in superelevation.

To compute for the Superelevation, four essential parameters are needed and these parameters are Gauge of the track (G), velocity of the body (v), radius of the curve (r) and acceleration due to gravity (g).

The formula for calculating the circular path motion:

S = Gv² / gr

Where:
S = Superelevation
G = Gauge of the track
v = velocity of the body
r = radius of the curve
g = acceleration due to gravity

Let’s solve an example;
Find the circular path motion when Gauge of the track is 14, velocity of the track is 47, radius of the curve is 21 and acceleration due to gravity is 9.8.

This implies that;
G = Gauge of the track = 14
v = velocity of the body = 47
r = radius of the curve = 21
g = acceleration due to gravity = 9.8

S = Gv² / gr
S = 14 x 47² / 9.8 x 21
S = 14 x 2209 / 205.8
S = 30926 / 205.8
S = 150.27

Therefore, the superelevation is 150.27 m.

Calculating the Gauge of the track when Superelevation, Velocity of the body, Radius of the curve and Acceleration due to gravity.

G = Sgr / v2

Where;
G = Gauge of the track
S = Superelevation
v = velocity of the body
r = radius of the curve
g = acceleration due to gravity

Let’s solve an example;
With a superelevation of 180, velocity of the body is 32, radius of the curve is 12 and acceleration due to gravity as 9.8. Find the gauge of the track?

This implies that;
S = Superelevation = 180
v = velocity of the body = 32
r = radius of the curve = 12
g = acceleration due to gravity = 9.8

G = Sgr / v2
G = 180 x 12 x 9.8 / 322
G = 21168 / 1024
G = 20.67

Therefore, the gauge of the track is 20.67.

Calculating the Velocity of the body when Superelevation, Gauge of the track, Radius of the curve and Acceleration due to gravity.

v = √Sgr / G

Where;
v = velocity of the body
S = Superelevation
G = Gauge of the track
r = radius of the curve
g = acceleration due to gravity

Let’s solve an example;
With a superelevation of 120, gauge of the track is 28, radius of the curve is 7 and acceleration due to gravity as 9.8. Find the velocity of the body?

This implies that;
S = Superelevation = 120
G = gauge of the track = 28
r = radius of the curve = 7
g = acceleration due to gravity = 9.8

v = √Sgr / G
v = √120 x 9.8 x 7 / 28
v = √8232 / 28
v = √294
v = 17.146

Therefore, the velocity of the body is 17.146.

Continue reading How to Calculate and Solve for Superelevation, Guage of Track, Velocity and Radius of a Body in Circular Path Motion | The Calculator Encyclopedia

How to Calculate and Solve for Mass, Velocity and Kinetic Energy | The Calculator Encyclopedia

The image above represents kinetic energy.

To compute for the kinetic energy, two essential parameters are needed and these parameters are mass (m) and velocity (v).

The formula for calculating the kinetic energy:

K.E. = 0.5mv²

Where;
K.E. = Kinetic Energy
m = Mass
v = Velocity

Let’s solve an example;
Find the kinetic energy when the mass is 6 and the velocity is 18.

This implies that;
m = Mass = 6
v = Velocity = 18

K.E. = 0.5mv²
K.E. = 0.5[6 x 18²]
K.E. = 0.5[6 x 324]
K.E. = 0.5[1944]
K.E. = 972

Therefore, the kinetic energy is 972 Joules (J).

Calculating the Mass when Kinetic Energy and Velocity is Given.

m = K.E / 0.5v2

Where;
m = Mass
K.E. = Kinetic Energy
v = Velocity

Let’s solve an example;
Find the mass when the kinetic energy is 320 and a velocity of 20.

This implies that;
K.E. = Kinetic Energy = 320
v = Velocity = 20

m = K.E / 0.5v2
m = 320 / 0.5 x 202
m = 320 / 0.5 x 40
m = 320 / 20
m = 16

Therefore, the mass is 16 kg.

Continue reading How to Calculate and Solve for Mass, Velocity and Kinetic Energy | The Calculator Encyclopedia