Thursday, April 7, 2011

QUESTION 1:
What is tire rotation?
ANSWER:
Tire rotation is the practice of swapping the front tires of a car with the back tires at regular intervals. The basic idea of tire rotation is to extend the life of the tires by allowing more even treadwear. Some specialty tire manufacturers do not recommend the practice for their customers, but most car owners benefit from a regular tire rotation. Some tire stores even include a basic tire rotation as part of their service contract.

QUESTION 2:

How  Propeller rotate converting rotational motion?

ANSWER:

A propeller is a type of fan which transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surfaces of the airfoil-shaped blade, and air or water is accelerated behind the blade. Propeller dynamics can be modeled by both Bernoulli's principle and Newton's third law. A propeller is often colloquially known as screw both in aviation and maritime.

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Thursday, March 17, 2011

ROTATIONAL MOTION

b)Application

Moment of inertia
  • Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion.
  • moment of inertia is also depend of axis of rotation and mass distribution of the body.
  • The point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses.
  • SI unit moments of inertia is kg m2   
  • It can be represented as I = m r2,  where I = moment of inertia kg m2 , m= mass (kg),    r = radius (m) 

(Question  with answer)
Question  1
A cable is wrapped around a uniform, solid cylinder of radius 'R' and mass 'M'. The cylinder rotates about its axis, and the cable unwinds without stretching or pulling. If the cable is pulled with a force of 'F' Newtons, what is its acceleration?
Hints
  1. What is the moment of inertia for a uniform, solid cylinder, with the axis through its center?
  2. What is the torque exerted?
  3. What is the relationship between acceleration of the cable, a, and the angular acceleration, ?

 
 

Diagram of the cable unwinding
from a cylinder.


Answer:
Drawing a diagram will aid us in solving this problem; refer to Diagram
  • For a uniform, solid cylinder of radius R and mass M, the moment of inertia is: 
  

The torque exerted by a force F is found to be: 
  

     = R F
since the force is perpendicular to the moment arm. (That is, =90, so sin(90)=1.) 
We also learned in this section that
     = I 

Solving for , we get:
 Question 2
A  rod of mass $m=5.3 {\rm kg}$ and length $l=1.3 {\rm m}$ rotates about a fixed frictionless
pivot located at one of its ends. The rod is released from rest at an angle $\theta=35^\circ$ beneath the horizontal. What is the angular acceleration of the rod immediately after it is released? 


\begin{figure*} \epsfysize =2.5in \centerline{\epsffile{rod.eps}} \end{figure*}

Answer: The moment of inertia of a rod of mass $m$ and length $l$ about an axis, perpendicular to its length, which passes through one of its ends is $I= (1/3) m l^2$.
                                             \begin{displaymath} I = \frac{5.3\times 1.3^2}{3} = 2.986 {\rm kg m^2}. \end{displaymath}
\begin{displaymath} I \alpha = \tau, \end{displaymath}


\begin{displaymath} x = \frac{l}{2} \cos\theta = \frac{1.3\times \cos 35^\circ}{2} = 0.532 {\rm m}. \end{displaymath}

\begin{displaymath} \tau = m g x. \end{displaymath}

\begin{displaymath} \alpha = \frac{\tau}{I} = \frac{m g x}{I} = \frac{5.3\times 9.81\times 0.532}{2.986} = 9.26 {\rm rad./s^2}. \end{displaymath}

Examples question:
Question 1:
A weight of mass m=4.8kg is suspended via a light inextensible cable which is wound around a pulley of mass M=13.5kg and radius b=0.8M. Treating the pulley as a uniform disk, find the downward acceleration of the weight and the tension in the cable. Assume that the cable does not slip with respect to the pulley.
\begin{figure*} \epsfysize =2.5in \centerline{\epsffile{pulw.eps}} \end{figure*}
Question 2:
weight of each stone has a mass of 23g and the throw distance of 8.2cm. when thrown into a place, with a speed of 31 m / s in 0.57s. assuming the weight and speed by throwing a stone that is thrown, determine the total torque is working on it.
 

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Thursday, February 17, 2011

name of group

NOOR AZILA BINTI ISMAIL
54136111059
NIK AZUANI BINTI NIK ALIAS
54136111091
NURUL SHUHADA BINTI CHE HASHIM
54136111067
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