
An aeroplane is flying with a uniform speed of 100 m/s along a circular path of radius 100 m. the angular speed of the aeroplane will be
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Answer
162.9k+ views
Hint:When a body is moving in a circular path then the position of the body along the circular path changes with time. As we know, the rate of change in the linear position of the body with respect to time is called the linear velocity. The rate of change of the angular position with respect to time is called the angular velocity.
Formula used:
\[v = \omega r\]
where v is the linear velocity, \[\omega \] is the angular velocity and r is the distance of the point from the axis of rotation.
Complete step by step solution:
It is given that the body is whirled in a horizontal circle. The radius of the circle is given as 100 m. The linear velocity of the body is given as 100 m/s. We need to find the angular velocity of any point on a circular path.

Image: Particle’s circular motion
Let the linear velocity of the body is \[\omega \]. The distance of the body from the axis of rotation, i.e. from the centre of the circular path is equal to the radius of the circular path.
\[r = 100\,m\]
\[\Rightarrow v = 100\,m/s\]
When a body is moving in a given circular path then the position of the body along the circular path changes with time. The rate of change in the linear position of the body with respect to time is called the linear velocity, v.
The linear velocity of the body is linearly related to the angular velocity of the body in the circular path as,
\[v = \omega r\]
Putting the values, we get the angular velocity of the body in circular orbit as,
\[\omega = \dfrac{v}{r}\]
\[\Rightarrow \omega = \left( {\dfrac{{100}}{{100}}} \right)rad/s\]
\[\therefore \omega = 1\,rad/s\]
Hence, the angular velocity of the body is equal to1 rad/s.
Therefore, the correct option is A.
Note: We should be careful while using the relation between the linear velocity and the angular velocity as it is for the instantaneous velocity. So, if we find the linear velocity of the accelerated body in a circular path then the obtained linear velocity will be corresponding to the angular velocity at that instant in time.
Formula used:
\[v = \omega r\]
where v is the linear velocity, \[\omega \] is the angular velocity and r is the distance of the point from the axis of rotation.
Complete step by step solution:
It is given that the body is whirled in a horizontal circle. The radius of the circle is given as 100 m. The linear velocity of the body is given as 100 m/s. We need to find the angular velocity of any point on a circular path.

Image: Particle’s circular motion
Let the linear velocity of the body is \[\omega \]. The distance of the body from the axis of rotation, i.e. from the centre of the circular path is equal to the radius of the circular path.
\[r = 100\,m\]
\[\Rightarrow v = 100\,m/s\]
When a body is moving in a given circular path then the position of the body along the circular path changes with time. The rate of change in the linear position of the body with respect to time is called the linear velocity, v.
The linear velocity of the body is linearly related to the angular velocity of the body in the circular path as,
\[v = \omega r\]
Putting the values, we get the angular velocity of the body in circular orbit as,
\[\omega = \dfrac{v}{r}\]
\[\Rightarrow \omega = \left( {\dfrac{{100}}{{100}}} \right)rad/s\]
\[\therefore \omega = 1\,rad/s\]
Hence, the angular velocity of the body is equal to1 rad/s.
Therefore, the correct option is A.
Note: We should be careful while using the relation between the linear velocity and the angular velocity as it is for the instantaneous velocity. So, if we find the linear velocity of the accelerated body in a circular path then the obtained linear velocity will be corresponding to the angular velocity at that instant in time.
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