Answer
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Hint: As a first step, you could find the time taken by the minute hand and second hand of the clock to complete one full rotation, that is, their respective time periods. Now you could recall the expression for angular speed in terms of the time period. Substituting the above found time period in this expression you will get the angular velocity of the minute hand and second hand and their difference will give you the answer.
Formula used:
The expression for angular velocity,
$\omega =\dfrac{2\pi }{T}rad/s$
Complete step by step solution:
In the question, we are asked to find the difference between the angular speed of the minute hand and the second hand of a clock.
In order to answer this question, let us find the time taken by the second hand and minute hand to complete a full rotation, that is, their respective time periods.
The time taken by the second hand to complete one full rotation, that is, the time period of the second hand will be,
${{T}_{s}}=60\sec $ ……………………………………….. (1)
The time taken by the minute hand to complete one full rotation, that is, the time period of the minute hand will be,
${{T}_{m}}=60\times 60s=3600s$ …………………………………….. (2)
Now, let us recall the expression for angular velocity which can be given by,
$\omega =\dfrac{2\pi }{T}rad/s$ ……………………………….. (3)
Now the angular velocity of second hand could be attained by substituting (1) in (3). So, the angular velocity of second hand will be,
${{\omega }_{s}}=\dfrac{2\pi }{{{T}_{s}}}$
$\Rightarrow {{\omega }_{s}}=\dfrac{2\pi }{60}$ …………………………….. (4)
Now we could substitute (2) in (3) to get the angular velocity of the minute hand. So, the angular velocity of the minute hand will be,
${{\omega }_{m}}=\dfrac{2\pi }{{{T}_{m}}}$
$\Rightarrow {{\omega }_{m}}=\dfrac{2\pi }{3600}$ …………………………………… (5)
For finding the difference in angular velocity we could subtract (5) from (4), that is, the required difference in angular speed is given by,
$\Delta \omega =\dfrac{2\pi }{60}-\dfrac{2\pi }{3600}$
$\Rightarrow \Delta \omega =\dfrac{2\pi }{60}\left( 1-\dfrac{1}{60} \right)$
$\Rightarrow \Delta \omega =\dfrac{\pi }{30}\left( \dfrac{59}{60} \right)$
$\therefore \Delta \omega =\dfrac{59\pi }{1800}rad/s$
Therefore, we found the difference between the angular speed of the minute hand and second hand of a clock to be$\dfrac{59\pi }{1800}rad/s$.
Note: The angular speed is defined as the ratio of angular distance covered to the time taken to cover that distance. The unit of angular speed is radians per second. The angular distance covered in one full rotation is $2\pi rad$ and hence the angular speed for one rotation is,
$\omega =\dfrac{2\pi }{t}$
The relation between the angular speed and linear speed is given by,
$v=R\omega $
Formula used:
The expression for angular velocity,
$\omega =\dfrac{2\pi }{T}rad/s$
Complete step by step solution:
In the question, we are asked to find the difference between the angular speed of the minute hand and the second hand of a clock.
In order to answer this question, let us find the time taken by the second hand and minute hand to complete a full rotation, that is, their respective time periods.
The time taken by the second hand to complete one full rotation, that is, the time period of the second hand will be,
${{T}_{s}}=60\sec $ ……………………………………….. (1)
The time taken by the minute hand to complete one full rotation, that is, the time period of the minute hand will be,
${{T}_{m}}=60\times 60s=3600s$ …………………………………….. (2)
Now, let us recall the expression for angular velocity which can be given by,
$\omega =\dfrac{2\pi }{T}rad/s$ ……………………………….. (3)
Now the angular velocity of second hand could be attained by substituting (1) in (3). So, the angular velocity of second hand will be,
${{\omega }_{s}}=\dfrac{2\pi }{{{T}_{s}}}$
$\Rightarrow {{\omega }_{s}}=\dfrac{2\pi }{60}$ …………………………….. (4)
Now we could substitute (2) in (3) to get the angular velocity of the minute hand. So, the angular velocity of the minute hand will be,
${{\omega }_{m}}=\dfrac{2\pi }{{{T}_{m}}}$
$\Rightarrow {{\omega }_{m}}=\dfrac{2\pi }{3600}$ …………………………………… (5)
For finding the difference in angular velocity we could subtract (5) from (4), that is, the required difference in angular speed is given by,
$\Delta \omega =\dfrac{2\pi }{60}-\dfrac{2\pi }{3600}$
$\Rightarrow \Delta \omega =\dfrac{2\pi }{60}\left( 1-\dfrac{1}{60} \right)$
$\Rightarrow \Delta \omega =\dfrac{\pi }{30}\left( \dfrac{59}{60} \right)$
$\therefore \Delta \omega =\dfrac{59\pi }{1800}rad/s$
Therefore, we found the difference between the angular speed of the minute hand and second hand of a clock to be$\dfrac{59\pi }{1800}rad/s$.
Note: The angular speed is defined as the ratio of angular distance covered to the time taken to cover that distance. The unit of angular speed is radians per second. The angular distance covered in one full rotation is $2\pi rad$ and hence the angular speed for one rotation is,
$\omega =\dfrac{2\pi }{t}$
The relation between the angular speed and linear speed is given by,
$v=R\omega $
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