Linear speed always measures the concrete distance traveled by a moving object. So, the linear speed measures in distance units per time units. For example meter per second. If an object is doing circular motion then the term linear is used as straightening out the arc traveled by the object alongside the circle. It results in a line of the same length. So, the common definition of speed as distance over time is correct. In this topic, we will discuss the linear speed formula with examples. Let us learn the concept!

**Linear Speed Formula**

**Concept of Linear Speed**

The linear speed of a point on a rotating object can be determined by its distance from the center of the rotation. The angular speed is measuring the angle that an object travels through in a certain amount of time. The angular speed will measure in units of radians per second i.e. radian per second.

It has 2\pi radians in a full circle. At a distance r i.e. radius from the center of the rotation. Then a point on the object has a linear speed equivalent to the angular speed multiplied by the distance r. Its measuring units are meters per second, meter per second.

**The Formula for Linear Speed**

We can compute Linear speed as,

\(s = \frac {d}{t} \)

Where,

s | Speed |

d | Distance |

t | Time |

\(linear speed = angular speed \times radius of the rotation\)

\(i.e. v = \omega \times r \)

v = linear speed (m per s)

\(\omega = angular speed (radians per s) \)

Where,

v | Linear speed |

\(\omega\) | Angular speed |

r | The radius of the rotation |

**Solved Examples**

Q.1: A power wheel begins and revolving at 10.0 revolutions per second. The diameter of the wheel is 4 m. What will be the linear speed of a point of the wheel on the surface, in meters per second?

Solution: First start to find the angular speed. The revolutions per second must be converted to radians per second. There are \(2\pi\) radians in a full circle.

\(\omega = 10.0 rev/s\)

\(r = \frac{4}{2} = 2 m\)

Using the formula

\(v = \omega \times r,\)

The linear speed of a point on the surface of the wheel is,

\(v = (62.8 radians/s) \times (2 m)\)

v = 125.6 m per s.

The linear speed of a point of the wheel is around 125.6 m per s.

Q.2: A sensor is connected under a car wheel that measures the linear speed. The sensor is 0.080 m from the center of rotation. At that point, the sensor reads that the linear speed of the wheel is around 8.00 m per s. So, if the radius of the wheel is 0.220 m, what is the linear speed on the outside edge of the wheel?

Solution: The linear speed is different at various distances from the center of rotation, but the angular speed is similar everywhere on the wheel. To solve this problem, the first step is to find the angular speed using the linear speed at the position of the sensor, 0.080 m.

The formula v = \(\omega \times r\) can be rearranged to solve for the angular speed:

\(v = \omega \times r\)

\(v = (100 rad/s) \times (0.220 m)\)

So,

v = 22.0 m

The linear speed of the car wheel at the outer edge is \(22.0 ms^{-1}\).

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