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RotationDynamics lab report

Rotationrefers to the movement of a body having the center in the circularpath and it moves within its rotation center.

Objectrotation includes the principles of translation motion. In rotatingbodies, the value of their angular velocity is calculated using theequation below.

ω= _{}

Wherev = velocity in circular path and is perpendicular to radius, r fromrotation center.

ω= angular velocity and it is a vector quantity.

Torquerefersto therateat which angularmomentumof a bodychangeswith timeinexplainingthedefinitionof torque,itis foundthateitherangularorinertial momentorbothcan change.Momentrefersto a mathematicalproductof massandforce,anditis usedto explainthetendencyof a singleormanyforcesthat cause(s)rotationof thebodyabout its axisof rotationAnexampleof therotating forceappliedto shaftcan causeaccelerationof thebodyat rest,anditleadsthemomentknownas torque

Angularacceleration (_{})is simply the rate at angular acceleration changes with time

_{}= _{}

UsingNewton’s second law the relationship that exists between torque andangular acceleration for the rotating motion is as shown below.

_{}= _{}

Where_{}= inertial moment

Materials

Thematerials are

Vernierdata collection interface

Loggerpro

Ringstand

Lightweight mass hanger

String

VernierMotion Rotary Center

VernierMotion Rotary Accessory kit

Balance

Drilledor slotted masses

Dataevaluation

Byconsidering the relationship between the net force and theacceleration of the object undergoing the translational motion,

_{}= _{}

Wherem is the mass of the object and a is the acceleration

Thefree body diagram is as shown below

DataProcessing

Thedata obtained above were processed in the results and analysis wherethe formula for calculating torque was derived.

Resultsand analysis

Radiusr = 60 cm = 0.6 m

Angularvelocity,ω = _{}

Mass,m = 0.15kg

Time/ s |
Cart speed v |
Angular velocity rad/s |

0 |
3 |
5.0 |

3 |
5 |
8.3 |

6 |
6 |
10.0 |

9 |
8 |
13.3 |

12 |
9 |
15.0 |

15 |
11 |
18.3 |

18 |
12 |
20.0 |

21 |
15 |
25.0 |

24 |
17 |
28.3 |

27 |
19 |
31.7 |

30 |
21 |
35.0 |

Slopeof the graph

=_{}

Angularacceleration,_{}= _{} =3.175rads/_{}

Theoreticalvalue of I = _{}

=_{}=0.027

Fromthe free body diagram

Forceacting on the body of mass, m = _{}

WhereT is the tension on the string

Torque_{}on the body = _{}

_{}=_{}

Fromthe equation above,

ButNewton’s 2^{nd}law _{}

Therefore,_{}

_{}= _{}= _{}

Wherer is the radius of circular path

_{}= _{}

From_{}=_{},

Impliesthat,

_{}= _{}

But_{}

So_{}= _{}

_{}= _{}

_{}= _{}

_{}= _{}

_{}= _{}

_{}= _{}

_{}= _{}= 0.29 Nm

Relationshipbetween torque and angular acceleration

Angularvelocity, ω= _{}

Angularacceleration of the gradient of the graph of angular velocity againsttime

Angularacceleration, _{}= _{}

Torquefrom the free body diagram above the tension T can calculated

From_{}

_{}

_{}= 0.995N

Torquein the string _{}= _{}

_{}= _{}= 0.597Nm

Conclusion

Intheanalysisof therelationshipbetween torqueandangularacceleration,itis seenthatthere is theexistenceof therelationbecausethebodyis rotatingin thecircularpathof radiusr wasseenin thederivationof theformula.Fromtheresultsobtainedfrom thenettorqueon thebodyis lessthan thetorqueon thestringbecausethestringwassubjectedto tension.Theexperimentalaimwassuccessfullyachievedwith theinvestigation