
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100s.
(a) What is the angular speed of motion?
(b) Is the acceleration vector constant? What is its magnitude?
Answer
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Hint: We need to understand the different physical quantities involved in the circular motion. The inter-dependence of these factors on each other can be used to find the solution required for this problem which are the speed and the acceleration.
Complete Step-by-Step Solution:
We are given that an insect is trapped in a groove which undergoes a circular motion completing each revolution in 100s which has a radius of 12 cm. We can easily find the angular speed of this system which the insect also is a part of.
The time period of revolution can be calculated as –
\[T=\dfrac{100}{7}s\]
We can find the angular speed using this time period as –
\[\begin{align}
& \omega =\dfrac{2\pi }{T} \\
& \Rightarrow \omega =\dfrac{2\times \dfrac{22}{7}}{\dfrac{100}{7}} \\
& \therefore \omega =0.44rad{{s}^{-1}} \\
\end{align}\]
This is the angular speed of the insect within the groove.
The motion has a constant speed throughout and is therefore a uniform circular motion. The velocity is always along the tangent to the circle at each point. The acceleration is due to the centripetal force which is along the radius of the circular motion. This direction changes at every instant of motion, i.e., the acceleration due to the motion is not a constant vector.
The magnitude can be given by the equation relating the angular speed and the radius of the circle as –
\[\begin{align}
& {{a}_{c}}={{\omega }^{2}}r \\
& \Rightarrow {{a}_{c}}=(0.44rad{{s}^{-1}}){{s}^{2}}\times (0.12m) \\
& \therefore {{a}_{c}}=0.023m{{s}^{-2}}=2.3cm{{s}^{-2}} \\
\end{align}\]
The solution for this problem is –
(a) \[\omega =0.44rad{{s}^{-1}}\]
(b) \[{{a}_{c}}=2.3cm{{s}^{-2}}\]
These are the required solutions.
Note:
The angular speed in a circular motion is the same as the angular frequency in a periodic motion as we can see in this problem. We should not get confused with these quantities in a normal periodic motion as they are not the same in such cases of motion.
Complete Step-by-Step Solution:
We are given that an insect is trapped in a groove which undergoes a circular motion completing each revolution in 100s which has a radius of 12 cm. We can easily find the angular speed of this system which the insect also is a part of.
The time period of revolution can be calculated as –
\[T=\dfrac{100}{7}s\]
We can find the angular speed using this time period as –
\[\begin{align}
& \omega =\dfrac{2\pi }{T} \\
& \Rightarrow \omega =\dfrac{2\times \dfrac{22}{7}}{\dfrac{100}{7}} \\
& \therefore \omega =0.44rad{{s}^{-1}} \\
\end{align}\]
This is the angular speed of the insect within the groove.
The motion has a constant speed throughout and is therefore a uniform circular motion. The velocity is always along the tangent to the circle at each point. The acceleration is due to the centripetal force which is along the radius of the circular motion. This direction changes at every instant of motion, i.e., the acceleration due to the motion is not a constant vector.
The magnitude can be given by the equation relating the angular speed and the radius of the circle as –
\[\begin{align}
& {{a}_{c}}={{\omega }^{2}}r \\
& \Rightarrow {{a}_{c}}=(0.44rad{{s}^{-1}}){{s}^{2}}\times (0.12m) \\
& \therefore {{a}_{c}}=0.023m{{s}^{-2}}=2.3cm{{s}^{-2}} \\
\end{align}\]
The solution for this problem is –
(a) \[\omega =0.44rad{{s}^{-1}}\]
(b) \[{{a}_{c}}=2.3cm{{s}^{-2}}\]
These are the required solutions.
Note:
The angular speed in a circular motion is the same as the angular frequency in a periodic motion as we can see in this problem. We should not get confused with these quantities in a normal periodic motion as they are not the same in such cases of motion.
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