Shutter Speed Examples: A Comprehensive Guide for Physics Students

shutter speed

Shutter speed is a fundamental concept in photography that determines the amount of time the camera’s shutter remains open, allowing light to reach the image sensor or film. This parameter plays a crucial role in capturing motion, controlling exposure, and minimizing camera shake. In this comprehensive guide, we will delve into the technical details, formulas, … Read more

The Inverse Relationship Between Torque and Speed: A Comprehensive Guide

relationship between torque and speed

The relationship between torque and speed is an essential concept in physics and mechanical engineering. As the speed of a rotating body increases, the torque it can produce decreases, and vice versa. This inverse relationship is crucial in understanding the behavior of various mechanical systems and designing efficient power transmission mechanisms.

Understanding the Torque-Speed Relationship

The relationship between torque and speed can be expressed mathematically as:

Torque = Power / Speed or τ = P / ω

Where:
P is the power (work done per unit time)
τ is the torque (rotational ability of a body)
ω is the angular speed/velocity (rate of change of angular displacement)

This formula can be rearranged to compute the angular velocity required to achieve a given torque and power:

Angular Velocity (ω) = Power (P) / Torque (τ)

The inverse relationship between torque and speed can be visualized in the following graph:

Torque-Speed Curve

As the graph shows, as the speed increases, the torque decreases, and vice versa. This relationship is crucial in the design and operation of various mechanical systems, such as electric motors, gearboxes, and power transmission systems.

Theoretical Foundations

relationship between torque and speed

The inverse relationship between torque and speed is rooted in the fundamental principles of physics, specifically the conservation of energy and the concept of power.

Conservation of Energy

The conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. In the case of a rotating body, the total energy is the sum of the kinetic energy (due to the body’s motion) and the potential energy (due to the body’s position).

As the speed of the rotating body increases, its kinetic energy increases, but its potential energy (in the form of torque) decreases. This inverse relationship between kinetic and potential energy is the basis for the inverse relationship between torque and speed.

Power and Torque

Power is defined as the rate of work done, or the amount of energy transferred per unit time. In the context of a rotating body, power can be expressed as the product of torque and angular velocity:

Power (P) = Torque (τ) × Angular Velocity (ω)

Rearranging this equation, we can derive the inverse relationship between torque and speed:

Torque (τ) = Power (P) / Angular Velocity (ω)

This equation clearly shows that as the angular velocity (speed) increases, the torque must decrease to maintain the same power output.

Applications and Examples

The inverse relationship between torque and speed has numerous applications in various fields, including:

Electric Motors

Electric motors are a prime example of the torque-speed relationship in action. Electric motors are designed to produce a specific torque-speed characteristic, which is determined by the motor’s construction and the applied voltage. As the load on the motor increases, the speed decreases, and the torque increases to maintain the same power output.

For instance, a 100-watt electric motor operating at 100 rad/s would have a torque of 1 Nm. If the angular velocity were to increase to 200 rad/s, the torque would decrease to 0.5 Nm, while the power output would remain the same.

Gearboxes

Gearboxes are used to transmit power from one rotating shaft to another, often with a change in speed and torque. The gear ratio of a gearbox is designed to match the torque-speed requirements of the driven load. By adjusting the gear ratio, the gearbox can increase the torque while decreasing the speed, or vice versa, depending on the application.

For example, a gearbox with a gear ratio of 1:5 would increase the torque by a factor of 5 while decreasing the speed by the same factor. This is useful in applications where high torque is required, such as in heavy machinery or industrial equipment.

Power Transmission Systems

The inverse relationship between torque and speed is also crucial in the design of power transmission systems, such as those found in vehicles, industrial machinery, and renewable energy systems. Engineers must carefully consider the torque-speed characteristics of the various components in the system to ensure efficient power transfer and optimal performance.

In a vehicle, for instance, the engine produces a specific torque-speed curve, which must be matched by the transmission and final drive gears to provide the desired acceleration and top speed. The gear ratios are selected to ensure that the engine operates within its optimal torque and speed range, maximizing the vehicle’s efficiency and performance.

Numerical Examples

Let’s consider a few numerical examples to illustrate the inverse relationship between torque and speed:

  1. Example 1: A motor has a power rating of 1 kW and is operating at an angular velocity of 100 rad/s. What is the torque produced by the motor?

Given:
– Power (P) = 1 kW = 1000 W
– Angular Velocity (ω) = 100 rad/s

Torque (τ) = Power (P) / Angular Velocity (ω)
Torque (τ) = 1000 W / 100 rad/s = 10 Nm

  1. Example 2: The same motor from Example 1 is now operating at an angular velocity of 200 rad/s. What is the new torque produced by the motor?

Given:
– Power (P) = 1000 W
– Angular Velocity (ω) = 200 rad/s

Torque (τ) = Power (P) / Angular Velocity (ω)
Torque (τ) = 1000 W / 200 rad/s = 5 Nm

As the angular velocity doubled, the torque decreased by a factor of 2, from 10 Nm to 5 Nm.

  1. Example 3: A gearbox has a gear ratio of 1:10. If the input shaft is rotating at 1000 rpm, what is the output shaft’s torque and speed?

Given:
– Gear Ratio = 1:10
– Input Shaft Speed = 1000 rpm

Output Shaft Speed = Input Shaft Speed / Gear Ratio
Output Shaft Speed = 1000 rpm / 10 = 100 rpm

Output Shaft Torque = Input Shaft Torque × Gear Ratio
Output Shaft Torque = Input Shaft Torque × 10

The gearbox has increased the torque by a factor of 10 while decreasing the speed by the same factor, as expected from the inverse relationship between torque and speed.

These examples demonstrate how the inverse relationship between torque and speed can be used to calculate the various parameters in mechanical systems, which is crucial for their design and optimization.

Conclusion

The inverse relationship between torque and speed is a fundamental concept in physics and mechanical engineering. This relationship is rooted in the principles of conservation of energy and the definition of power, and it has numerous applications in various fields, such as electric motors, gearboxes, and power transmission systems.

By understanding the mathematical relationship between torque and speed, engineers can design and optimize mechanical systems to achieve the desired performance characteristics. The numerical examples provided in this article illustrate how this relationship can be applied to solve real-world problems and ensure the efficient operation of rotating machinery.

References

  1. Vedantu. (n.d.). Relation between Torque and Speed. Retrieved from https://www.vedantu.com/physics/relation-between-torque-and-speed
  2. Byjus. (n.d.). Relation between Torque and Speed. Retrieved from https://byjus.com/physics/relation-between-torque-and-speed/
  3. I Need Motors. (n.d.). What is the Relationship Between Speed and Torque? Retrieved from https://www.ineedmotors.com/news/what-is-the-relationship-between-speed-and-tor-53813295.html

Uniform vs Non-Uniform Speed: A Comprehensive Guide for Physics Students

uniform vs non uniform speed

Uniform motion and non-uniform motion are two fundamental concepts in physics that describe the movement of objects. Understanding the differences between uniform and non-uniform speed is crucial for analyzing and predicting the motion of objects in various real-world scenarios. This comprehensive guide will provide you with a deep dive into the technical details, formulas, examples, … Read more

Is Horizontal Speed Constant: Why, How, When, Problems

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The article discusses about is horizontal speed constant, along with its problem examples.

The horizontal speed is constant throughout the path as no force acts on the projectile in the horizontal direction to alter its acceleration. At the same time, the projectile travels downwards due to the gravity force, which defines the projectile motion along with constant horizontal speed. 

Read more about Constant Horizontal Velocity.

The projectile travels vertically and horizontally at the exact time. Firstly, it happens because the air resistance force or air drag on the projectile is neglected as it is minimal. Hence, there is no force acting on the projectile in the horizontal direction to change its horizontal component of speed. As a result, the projectile traveled a constant horizontal distance (x) in unit time with constant horizontal speed (vH).

Is Horizontal Speed Constant
Is Horizontal Speed Constant

x = vHt …………. (*)

The horizontal speed equals the initial velocity u at the projectile launch. Hence,

vH = u …………… (1)

Read more about Speed.

On the other hand, the gravity force, a non-contact force, always acts on the projectile after it is launched, changing its vertical speed component. That’s why the projectile accelerated at 9.8m/s2 downward, which changes its vertical speed. The vertical acceleration is the change in vertical speed (vV-uV) per unit time.

Projectile Motion due to Gravity
Vertical Motion due to Gravity

a =vV – uV/t

If we rearrange, we can obtain the kinematics equation of motion.

vV = uV + at ……………. (2)

Since the vertical component of speed uV is zero initially and acceleration(g) is driven by the gravity force,

vV = gt …………… (3)

If a cannonball fired from the cannon horizontally from the cliff, it takes 5 sec to reach the ground. Also, it falls on the ground at 20m from the base of the cliff. Calculate both the horizontal and vertical speeds of the cannonball after the fire? 

Given:

t = 5s

x = 20m

g = 9.8m/s2

To Find:

  1. vH =?
  2. vV =?

Formula:

  1. x = vHt
  2. vV = uV + gt

Solution:

The horizontal speed is calculated as,

x = vHt

vH = x/t

Substituting all values,

vH = 20 / 5

vH = 4

The horizontal speed of the cannonball is 4m/s

The vertical speed is calculated from the kinematics equation of motion,

vV = uV + gt

Since initial vertical velocity is zero. i.e., uV = 0.

Substituting all values,

vV = 0 + 9.8 x 5

vV = 49

The vertical speed of the cannonball is 49m/s.

Read more about How to Calculate Momentum.

Why is Horizontal Speed Constant?

The horizontal speed is constant in projectile motion due to no forces in the horizontal direction. 

The horizontal speed implies that the projectile moves in a straight path without changing its direction. No net forces are acting in a horizontal direction after the projectile is launched. Hence, the projectile’s horizontal speed stays constant throughout the trajectory motion due to zero acceleration. 

Read more about Net Forces.

In the previous article, we comprehended the difference between horizontal speed and horizontal velocity. In projectile motion, the velocity has horizontal and vertical components. The vertical velocity emerges due to gravity force that acts downward on the projectile.

In contrast, no forces act horizontally on the projectile, guiding the projectile to move with constant horizontal velocity. That’s the reason the projectile moves with both vertical and horizontal velocity at the same time after it is launched. 

But just after launching the projectile, its vertical velocity is zero initially. Hence, the projectile initially moves with the constant horizontal speed in a linear path. Once the vertical velocity increases, changing the projectile’s direction, its horizontal speed becomes the horizontal velocity. 

The conversion from horizontal speed to horizontal velocity depends on the launch angle of the projectile. Lesser the launch angle with horizontal, the projectile moves with constant horizontal speed more than horizontal velocity. If the launch angle increases, the horizontal speed becomes the horizontal velocity with an increase in vertical velocity. 

launch angle in projectile motion
Launch Angles in Projectile Motion (credit: shuttertstock)

Suppose a cannonball is a fire from a cannon located at the cliff’s edge. If we launch the cannonball at an angle of 0°, it moves in a straight direction with constant horizontal speed unless an external force stops it. Suppose we increase the launch angle to 30°, then its horizontal speed becomes horizontal velocity due to an increase in vertical velocity, and the cannonball starts to descend towards the ground. 

Horizontal Speed for various Launch Angles
Horizontal Speed for Various Launch Angles

If the launch angle becomes 80°, then the cannonball moves with constant horizontal velocity, and it initially gets vertical velocity immediately after it launches. The cannonball moves upwards with both changing vertical velocity and constant horizontal velocity.

At maximum height, the vertical velocity becomes zero, at which the cannonball moves only the constant horizontal velocity. Below the maximum height, it gains the vertical velocity as it starts to move towards the ground, and hence, its acceleration is in the same direction as the gravity force. 

Read more about Gravitational Acceleration.

A football moves 20m along the ground at 10 sec after a player kicks it. When a player kicks the football at 60°, it changes velocity from 5m/s to 10m/s; it reaches the ground in 5 sec after moving in a parabolic path. Calculate the horizontal speed and vertical velocity in both cases. 

Given:

Case 1:

x = 20m

t = 10sec

Case 2:

θ = 60°

u = 5m/s

uV = 10m/s

t = 5sec

g = 9.8m/s2

To Find:

  1. vH =?
  2. vV =?

Formula:

  1. x = vHt
  2. vV = uV + gt

Solution:

1) As the football moves along the ground, the launch angle θ is zero.

Hence, in case 1, the football moves with constant horizontal speed, and the vertical velocity vV is zero.

The horizontal speed of football is calculated as,

x = vHt

vH = x/t

Substituting all values,

vH = 20/10

vH = 2

The football moves along the ground with a constant horizontal speed of 2m/s.

2) As the football launches at 60°, instead of horizontal speed, the football moves with a horizontal velocity equal to the initial velocity of football u = 5m/s.

The vertical velocity of the football is calculated as,

vV = uV + gt

Substituting all values,

vV = 10 + 9.8 x 5

vV = 10 + 49

vV = 59

The vertical velocity of the football launched at 60° is 59m/s.

How is Horizontal Speed Constant?

The horizontal speed is constant only when the object moves in a parabolic path.

The projectile motion in parabolic trajectory defined by horizontal and vertical velocity is only controlled by the downward gravity force, which sets up only vertical acceleration. Since there is no net force to set up the horizontal acceleration, the projectile moves with constant horizontal speed.  

Read more about Relative Motion.

Earlier, we discussed the launch angle and horizontal speed relation. We will learn more about the time in which the projectile remains in the air. The best combination for the time in the air when the projectile moves with horizontal speed is launching at 45°, which provides the maximum horizontal distance to the projectile. The basketball and javelin thrower is a projectile motion example that proves this sentence true. 

Horizontal Speed in Basketball
Horizontal Speed in Basketball
(credit: shutterstock)

We have determined earlier the formula about horizontal distance x. The vertical distance travelled by projectile is calculated using second kinematics equation of motion

y = uyt + 1/2 gt2

But uV is zero. 

y = gt2/2 ……….. (4)

Substituting value of t from equation (*), t = x/vx , we get

y = g(x/vx)2 / 2

y = gx/2vx2 ……………… (5)

The above equation is the parabola equationwhich predicts how much the projectile travels vertical and horizontal after launch. It shows the parabolic trajectory of the projectile, which launched parallel to horizontal. 

Let’s calculate the magnitude of the resultant velocity of the projectile using the Pythagoras theorem

v2 =vx2 + vy2

From equation (1) and (4).

v= √(u2+g2t2) ……………. (6)

Now let’s calculate the formula about how the projectile remains in the air from the second kinematics equation of motion for vertical distance traveled (4), 

t = √(2y/g) ………… (7)

That means, if we determine the time for a projectile in the air (t), we can also calculate the horizontal distance travelled (x) as the constant horizontal speed (vH) is the same as the initial velocity at which the projectile launched during the trajectory.

Therefore, the horizontal distance (x) by the projectile with constant horizontal speed (vH) is given by,

x = vHt

Using equation (1) and (7), we get

x = u √(2y/g) ………………(8)

Horizontal and Vertical Distance in Projectile Motion
Horizontal and Vertical Distance in Projectile Motion
(credit: shuttertstock)

Read more about How to Calculate Distance.

Suppose a basketball is thrown at angle 30°, which moves initially at 5m/s and then reaches the basket in 4sec after parabolic trajectory. 

Calculate the vertical distance travelled by the basketball.

Calculate the horizontal distance travelled by the basketball.

Calculate the horizontal speed of the basketball.

Calculate the resultant velocity of the basketball.

Given:

θ = 30°

u = 5m/s

t = 4s

g = 9.8m/s2

To Find:

  1. y =?
  2. x =?
  3. vH =?
  4. v =?

Formula:

  • y = gt2/2
  • x = u √(2y/g)
  • x = vHt
  • v = √(u2+g2t2)

Solution:

1)The vertical distance travelled by the basketball is calculated as,

y =gt2/2

Substituting all values,

y = 9.8*42/2

y = 78.4

The vertical distance travelled by basketball, launched at 30°, is 78.4m.

2) The horizontal distance travelled by the basketball is calculated as,

x = u √(2y/g)

Substituting all values,

x = 5 √(2*74.8/9.8)

x = 5√15.26

x = 5×3.9

x = 19.5

The horizontal distance travelled by basketball, launched at 30°, is 19.5m.

3)The horizontal speed of basketball is calculated as,

x = vHt

vH = x/t

Substituting all values,

vH = 19.5/4

vH = 4.87

The basketball moves with a constant horizontal speed of 4.87m/s.

4)The resultant velocity of basketball is calculated as,

v = √(u2+g2t2)

Substituting all values,

v = √(52+9.82*42)

v = √(25+1536.64)

v = √1561.64

v = 39.51

The resultant velocity of a basketball, launched at 30°, is 39.51m/s.


Also Read:

Instantaneous vs Average Speed: A Comprehensive Guide for Physics Students

instantaneous vs average speed

Instantaneous speed and average speed are two fundamental concepts in physics that describe the motion of objects. These two measures of speed provide different insights into the movement of an object, and understanding the distinction between them is crucial for physics students. In this comprehensive guide, we will delve into the details of instantaneous and … Read more