
What is the angular speed of the minute hand of a clock? If the minute hand is 5 cm long. What is the linear speed of its tip?
Answer
499.8k+ views
Hint: To solve this question students should know what is angular speed, linear speed and the formulas for the same. Angular speed is the rate of change of the angle of the radian of an object. And the linear speed is defined as the rate of change of distance covered.
Complete step by step solution:
The clock is a circular path in which the minute hand rotates in a clockwise direction. Let the minute hand of the clock be of length L and take that length as the radius of the circular path of the clock in which the minute hand rotates.
As we know the angular speed is the rate of change of angular displacement.
Angular speed $\left( \omega \right) = \dfrac{{angular{\text{ displacement}}\left( \theta \right)}}{{time\left( t \right)}}$
And the linear speed is the rate of change of total distance travelled.
Linear speed $\left( v \right) = radius(r) \times angular{\text{ speed}}(\omega )$
In a complete revolution, the minute hand of the clock covers a distance of ${360^0}$ or $2\pi $radian in one hour that is $3600$ seconds.
So, we have angular displacement $\left( \theta \right) = 2\pi {\text{ radian}}$and time $\left( t \right) = 3600{\text{ seconds}}$
We know that,
$angular{\text{ speed}}\left( \omega \right) = \dfrac{{angular{\text{ displacement}}\left( \theta \right)}}{{time\left( t \right)}}$
Substituting the values, we get
$angular{\text{ speed}}\left( \omega \right) = \dfrac{{2\pi }}{{3600}}$
$\therefore angular{\text{ speed}}\left( \omega \right) = 1.74 \times {10^{ - 3}}{\text{ rad/s}}$
We have length of the minute hand as $5{\text{ cm}}$convert it into meters as the standard unit of the distance is meters.
So,
$5{\text{ cm = 5 }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ m}}$
Therefore, from the equation of linear speed we have
$linear{\text{ speed}}\left( v \right) = radius(r) \times angular{\text{ speed}}(\omega )$
Now substituting the values,
$linear{\text{ speed}}\left( v \right) = 5 \times {10^{ - 2}} \times 1.74 \times {10^{ - 3}}$
Now further simplifying, we get
$linear{\text{ speed}}\left( v \right) = 8.7 \times {10^{ - 5}}{\text{ m/s}}$
Note: SI unit of angular speed is rad/sec but in some of the cases it is given as rpm that is revolution per minute. Revolution per minute is not a SI unit it’s just a semantic annotation rather than a unit. The number of turns in one minute is known as rpm (revolution per minute). $1{\text{ rad/s = }}\dfrac{1}{{2\pi }}hz{\text{ = }}\dfrac{{60}}{{2\pi }}rpm$ . Students should be careful that all units of the given physical quantity should be in the same units.
Complete step by step solution:
The clock is a circular path in which the minute hand rotates in a clockwise direction. Let the minute hand of the clock be of length L and take that length as the radius of the circular path of the clock in which the minute hand rotates.
As we know the angular speed is the rate of change of angular displacement.
Angular speed $\left( \omega \right) = \dfrac{{angular{\text{ displacement}}\left( \theta \right)}}{{time\left( t \right)}}$
And the linear speed is the rate of change of total distance travelled.
Linear speed $\left( v \right) = radius(r) \times angular{\text{ speed}}(\omega )$
In a complete revolution, the minute hand of the clock covers a distance of ${360^0}$ or $2\pi $radian in one hour that is $3600$ seconds.
So, we have angular displacement $\left( \theta \right) = 2\pi {\text{ radian}}$and time $\left( t \right) = 3600{\text{ seconds}}$
We know that,
$angular{\text{ speed}}\left( \omega \right) = \dfrac{{angular{\text{ displacement}}\left( \theta \right)}}{{time\left( t \right)}}$
Substituting the values, we get
$angular{\text{ speed}}\left( \omega \right) = \dfrac{{2\pi }}{{3600}}$
$\therefore angular{\text{ speed}}\left( \omega \right) = 1.74 \times {10^{ - 3}}{\text{ rad/s}}$
We have length of the minute hand as $5{\text{ cm}}$convert it into meters as the standard unit of the distance is meters.
So,
$5{\text{ cm = 5 }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ m}}$
Therefore, from the equation of linear speed we have
$linear{\text{ speed}}\left( v \right) = radius(r) \times angular{\text{ speed}}(\omega )$
Now substituting the values,
$linear{\text{ speed}}\left( v \right) = 5 \times {10^{ - 2}} \times 1.74 \times {10^{ - 3}}$
Now further simplifying, we get
$linear{\text{ speed}}\left( v \right) = 8.7 \times {10^{ - 5}}{\text{ m/s}}$
Note: SI unit of angular speed is rad/sec but in some of the cases it is given as rpm that is revolution per minute. Revolution per minute is not a SI unit it’s just a semantic annotation rather than a unit. The number of turns in one minute is known as rpm (revolution per minute). $1{\text{ rad/s = }}\dfrac{1}{{2\pi }}hz{\text{ = }}\dfrac{{60}}{{2\pi }}rpm$ . Students should be careful that all units of the given physical quantity should be in the same units.
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