# Rotational Kinematics Numerical Problems and solutions

Last updated on April 12th, 2021 at 09:27 am

This post is all about **Rotational Kinematics Numerical Problems and solutions**. We will use the following four rotational kinematic equations (presented together with their translational counterparts) to solve the numerical problems.

## Rotational kinematic equations | equations used to solve rotational motion numerical

In these equations, the subscript 0 denotes initial values (*ω_{0 }*and

*are initial values). The 4 equations in the first column represent the rotational kinematics. The second column consists of the translational equivalents.*

*v*_{o}Rotational Kinetic equations | Translational kinetic equations | Constants |
---|---|---|

θ=(1/2)( + ω_{0}ω)t | s=(1/2)( + v_{o}v)t | α, a |

ω = ω_{0 }+ αt | v = v_{o} + at | α, a |

θ=ωt+(1/2)αt_{0}^{2} | s=vt+(1/2)at_{0}^{2} | α, a |

ω^{2} = ω_{0}^{2}+ 2αθ | v = ^{2}v_{o}^{2} + 2as | α, a |

**four rotational kinematic equations (presented together with their translational counterparts)**

## Numerical problem on Angular Displacement | Rotational Kinematics Numerical

**1. A rigid body turns through 1.85 radians. Express this in degrees and revolutions**.

**Solution:**Using the fact that π radians equal 180 deg, then

1.85 rad = (1.85 rad)(180 deg /π rad) = 106 deg ……….(1)

Then, using the fact that 1 revolution = 2π rad to get

1.85 rad = (1.85 rad)(1 rev / 2π rad) = 0.294 rev ……………… (2)

## Rotational Kinematics Numerical on Angular Velocity and Acceleration

**2. Long ago people listened to music which was stored on “phonograph records”. These records turned at a rate of 33.3 revolutions per minute. Express this rotation rate in radians per second.**

**Solution:**

Using the relations **1 rev = 2π rad** as well as **1 min = 60 s** to convert the units:

33.3 rpm = 33.3 rev/min = (33.3 x 2π ) rad / 60 secs = 3.49 rad/s

## Numerical problem on Rotational Motion with Constant Angular Acceleration

**3. A flywheel has a constant angular deceleration of 2.0 rad/ s^{2}. (a) Find the angle through which the flywheel turns as it comes to rest from an angular speed of 220 rad/s. (b) Find the time required for the flywheel to come to rest.**

**Solution**(a)

α = – **2.0 rad/s**^{2}

**ω**_{0}** = ****220 rad/s**

**ω**** = 0**

**θ ****= ?**

Using this rotational kinematics equation: **ω**^{2}** = ω**_{0}^{2}**+ 2αθ**

=> **0 = 220**^{2}**+ 2.(-2).θ**

=> **θ = (220 x 220) / 4 = 12100 rad = 1.21 x 10**^{4}** rad**

**Solution (b)**

α = – **2.0 rad/s**^{2}

**ω**_{0}** = ****220 rad/s**

**ω**** = 0**

**t = ?**

**ω= ω**_{0}**+ α****t**

**0 = 220 + (-2)t**

**t = 110 sec**

## Rotational Kinematics Numerical based on the Relation Between Angular and Linear Quantities

**4. A string trimmer is a tool for cutting grass and weeds. It utilizes a length of nylon “string” that rotates about an axis perpendicular to one end of the string. The string rotates at an angular speed of 47 rev/s, and its tip has a tangential speedof 54 m/s. What is the length of the rotating string?**

**Solution:**

The rotating string is shown in Fig. 1. The angular velocity of the string’s rotation is**ω**= 2**π**f = 2**π**(47) rad/s = 295 rad/s

We have the speed of the tip of the string (which is a distance r from the axis), and this is the linear speed v.

v= 54 m/s

Now from Equation v=**ω**r, we can get the length of the string, r:

r =v/**ω** = 54/295 m = 0.183 m

So the string is 0.183 m long.

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