
The linear speed of a particle moving in a circle of radius R varies with time t as $V={{V}_{0}}-kt$, where k is a positive constant and ${{v}_{0}}$is initial velocity. At what time t, the angular velocity becomes equal to the angular acceleration?
A. $t=\dfrac{{{v}_{0}}-k}{2k}$
B. $t=\dfrac{{{v}_{0}}-k}{k}$
C. $t=\dfrac{{{v}_{0}}-k}{4k}$
D. $t=\dfrac{{{v}_{0}}-2k}{k}$
Answer
415.5k+ views
Hint: As the very first step, one could read the question well and hence note down the important points from it. Now recall the expressions for the given quantities in terms of the t that has to be found. Now equate them and hence get the required time.
Formula used:
Angular velocity,
$\omega =\dfrac{v}{R}=\dfrac{{{v}_{0}}-kt}{R}$
Angular acceleration,
$\alpha =\dfrac{d\omega }{dt}=-\dfrac{k}{R}$
Complete step by step solution:
In the question, the linear speed of a particle moving around in a circle of radius R is varying with time t as given as following,
$V={{V}_{0}}-kt$
Here k is a positive constant and ${{v}_{0}}$is the initial velocity.
Now we are supposed to find the time t at which the angular velocity becomes equal to the angular acceleration.
In order to answer this, we need to find the expression for the mentioned physical quantities.
For angular velocity we have the relation given by,
$\omega =\dfrac{v}{R}=\dfrac{{{v}_{0}}-kt}{R}$………………………………………….. (1)
Now, we have the expression for angular acceleration given by,
$\alpha =\dfrac{d\omega }{dt}=-\dfrac{k}{R}$…………………………………………… (2)
Now, let us assume that the magnitudes of these angular velocity and angular acceleration to be equal as per the given question,
$\left| \omega \right|=\left| \alpha \right|$
$\Rightarrow \dfrac{{{v}_{0}}-kt}{R}=\dfrac{k}{R}$
$\therefore t=\dfrac{{{v}_{0}}-k}{k}$
Therefore, we found the time at which the angular acceleration becomes equal to the angular velocity to be $t=\dfrac{{{v}_{0}}-k}{k}$. Option B is found to be the correct answer.
Note: Here, we have used the reverse mechanism. That is, we have assumed the magnitudes of the angular velocity and angular acceleration to be equal and hence have derived the required time from this assumption. So basically we have found the answer from the condition given in the question that has to be met.
Formula used:
Angular velocity,
$\omega =\dfrac{v}{R}=\dfrac{{{v}_{0}}-kt}{R}$
Angular acceleration,
$\alpha =\dfrac{d\omega }{dt}=-\dfrac{k}{R}$
Complete step by step solution:
In the question, the linear speed of a particle moving around in a circle of radius R is varying with time t as given as following,
$V={{V}_{0}}-kt$
Here k is a positive constant and ${{v}_{0}}$is the initial velocity.
Now we are supposed to find the time t at which the angular velocity becomes equal to the angular acceleration.
In order to answer this, we need to find the expression for the mentioned physical quantities.
For angular velocity we have the relation given by,
$\omega =\dfrac{v}{R}=\dfrac{{{v}_{0}}-kt}{R}$………………………………………….. (1)
Now, we have the expression for angular acceleration given by,
$\alpha =\dfrac{d\omega }{dt}=-\dfrac{k}{R}$…………………………………………… (2)
Now, let us assume that the magnitudes of these angular velocity and angular acceleration to be equal as per the given question,
$\left| \omega \right|=\left| \alpha \right|$
$\Rightarrow \dfrac{{{v}_{0}}-kt}{R}=\dfrac{k}{R}$
$\therefore t=\dfrac{{{v}_{0}}-k}{k}$
Therefore, we found the time at which the angular acceleration becomes equal to the angular velocity to be $t=\dfrac{{{v}_{0}}-k}{k}$. Option B is found to be the correct answer.
Note: Here, we have used the reverse mechanism. That is, we have assumed the magnitudes of the angular velocity and angular acceleration to be equal and hence have derived the required time from this assumption. So basically we have found the answer from the condition given in the question that has to be met.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
