Answer
Verified
329k+ views
Hint: Motion in a circular path means that the angular velocity of the particle undergoing circular motion is constant with time. One should always take care of the difference between scalar and vector quantity. Speed is a scalar quantity and velocity is a vector quantity. If the speed is constant, it doesn’t imply that the acceleration is zero but if the velocity is constant, it always implies that acceleration is zero or motion is non-accelerated.
Formula used:
$a= \dfrac{dv}{dt}$,$v=R \omega$
Complete answer:
To understand the statement mathematically, let’s consider a particle undergoing circular motion with speed ‘v’ and radius ‘R’ on a circular track.
Hence, its angular velocity ‘$\omega$’ will be $\omega = \dfrac{v}{R}$.
Let’s see properly marked diagram from top:
Resolving the speed as velocity into x and y components, so: $\vec { v } =\quad (vcos\omega t\hat { )i } -(vsin\omega t)\hat { j }$
To find the acceleration of the particle, let’s differentiate the velocity expression.
$a= \dfrac{dv}{dt}$
$a= \dfrac{d}{dt}\left ((vcos\omega t\hat { )i} -(vsin\omega t)\hat { j }\right )$
=$v \omega (-sin \omega t) \hat i - v \omega (cos \omega t) \hat j$ $\left ( \dfrac{d}{dt} cos(at+v) =-asin(at+v), \dfrac{d}{dt} sin(at+v) =acos(at+v) \right )$
= $-v \omega (sin \omega t \ \hat i+cos \omega t \ \hat j)$
Putting $v=R \omega$, we get
$\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$
Hence, vector form of centripetal acceleration is: $\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$ which shows that it’s direction is towards the centre and along the string at every point of circular track.
Expression for centripetal acceleration of a body undergoing circular motion:
As we derived $\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$
To find the magnitude:
$|\vec a| = \omega R^2 \sqrt{(sin \omega t)^2 + (cos \omega t)^2}$=$\omega R^2$ [ $sin^2 \theta + cos^2 \theta = 1$]
Hence centripetal acceleration is $\omega R^2$.
Note:
As we derived the expression for centripetal acceleration of the body undergoing uniform circular motion, we can see that the magnitude of this acceleration is constant with time. One should always remember that whenever a body moves in a curve (basically changes its direction of motion), it will always experience centripetal force acting on it. In short, direction can’t be changed without the application of force. In the above case, the string is providing necessary centripetal force.
Formula used:
$a= \dfrac{dv}{dt}$,$v=R \omega$
Complete answer:
To understand the statement mathematically, let’s consider a particle undergoing circular motion with speed ‘v’ and radius ‘R’ on a circular track.
Hence, its angular velocity ‘$\omega$’ will be $\omega = \dfrac{v}{R}$.
Let’s see properly marked diagram from top:
Resolving the speed as velocity into x and y components, so: $\vec { v } =\quad (vcos\omega t\hat { )i } -(vsin\omega t)\hat { j }$
To find the acceleration of the particle, let’s differentiate the velocity expression.
$a= \dfrac{dv}{dt}$
$a= \dfrac{d}{dt}\left ((vcos\omega t\hat { )i} -(vsin\omega t)\hat { j }\right )$
=$v \omega (-sin \omega t) \hat i - v \omega (cos \omega t) \hat j$ $\left ( \dfrac{d}{dt} cos(at+v) =-asin(at+v), \dfrac{d}{dt} sin(at+v) =acos(at+v) \right )$
= $-v \omega (sin \omega t \ \hat i+cos \omega t \ \hat j)$
Putting $v=R \omega$, we get
$\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$
Hence, vector form of centripetal acceleration is: $\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$ which shows that it’s direction is towards the centre and along the string at every point of circular track.
Expression for centripetal acceleration of a body undergoing circular motion:
As we derived $\vec a = -\omega R^2 ( sin \omega t\ \hat i + cos \omega t\ \hat j)$
To find the magnitude:
$|\vec a| = \omega R^2 \sqrt{(sin \omega t)^2 + (cos \omega t)^2}$=$\omega R^2$ [ $sin^2 \theta + cos^2 \theta = 1$]
Hence centripetal acceleration is $\omega R^2$.
Note:
As we derived the expression for centripetal acceleration of the body undergoing uniform circular motion, we can see that the magnitude of this acceleration is constant with time. One should always remember that whenever a body moves in a curve (basically changes its direction of motion), it will always experience centripetal force acting on it. In short, direction can’t be changed without the application of force. In the above case, the string is providing necessary centripetal force.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE