
The velocity and acceleration vectors of a particle undergoing circular motion are \[\vec v = 2{\hat i_{}}m/s\]and \[\vec a = 2\hat i + 4{\hat j_{}}m/{s^2}\] respectively at an instant of time. The radius of the circle is
(A) 1m
(B) 2m
(C) 3m
(D) 4m
Answer
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Hint It is given that the particle undergoing circular motion has velocity and acceleration as components of x and y axes. We know that velocity is with respect to the x axes and acceleration is having two components tangential and radial. Find the radius value using the radial component.
Complete Step By Step Solution
We have a particle that is moving in a circular motion of radius r, with a velocity of \[\vec v = 2{\hat i_{}}m/s\], which is a vector quantity. Now since the velocity is given only in one component , the velocity is towards the x axis of the reference frame. Velocity is tangential to the motion of the particle, hence it is also called tangential velocity.
In general physics, in a circular motion, the acceleration of the particle is said to be at the centre and should be perpendicular to the direction of velocity. The acceleration is centripetal since the velocity is said to be tangential. Now, looking at the vector representation of the acceleration of the particle, we can say the acceleration is normal to velocity.
Now acceleration consists of 2 components , tangential component and radial component. The tangential component of the acceleration of the particle is similar to that of the tangential velocity and hence it is assumed that it represents the same. However, the radial component now describes the radius of the particle it travelled through the circular path.
\[ \Rightarrow \vec a = 2\hat i + 4{\hat j_{}}m/{s^2}\],\[{a_x} = 2\] and \[{a_y} = 4\]
Now formula to calculate radial acceleration in a centripetal acceleration is given as ,
\[{a_r} = \dfrac{{{v^2}}}{r}\]
\[ \Rightarrow 4 = \dfrac{{{2^2}}}{r}\](Since, \[{a_y} = 4\] and \[\vec v = 2\])
\[ \Rightarrow r = \dfrac{4}{4} = 1m\]
Hence the radius of the circle is 1m.
Thus, option (a) is the right answer for the given question.
Note In a centripetal acceleration of the body, the acceleration which is directed along the radius of the path is called radial acceleration. In this case, the tangential velocity is said to be normal to the acceleration.
Complete Step By Step Solution
We have a particle that is moving in a circular motion of radius r, with a velocity of \[\vec v = 2{\hat i_{}}m/s\], which is a vector quantity. Now since the velocity is given only in one component , the velocity is towards the x axis of the reference frame. Velocity is tangential to the motion of the particle, hence it is also called tangential velocity.
In general physics, in a circular motion, the acceleration of the particle is said to be at the centre and should be perpendicular to the direction of velocity. The acceleration is centripetal since the velocity is said to be tangential. Now, looking at the vector representation of the acceleration of the particle, we can say the acceleration is normal to velocity.
Now acceleration consists of 2 components , tangential component and radial component. The tangential component of the acceleration of the particle is similar to that of the tangential velocity and hence it is assumed that it represents the same. However, the radial component now describes the radius of the particle it travelled through the circular path.
\[ \Rightarrow \vec a = 2\hat i + 4{\hat j_{}}m/{s^2}\],\[{a_x} = 2\] and \[{a_y} = 4\]
Now formula to calculate radial acceleration in a centripetal acceleration is given as ,
\[{a_r} = \dfrac{{{v^2}}}{r}\]
\[ \Rightarrow 4 = \dfrac{{{2^2}}}{r}\](Since, \[{a_y} = 4\] and \[\vec v = 2\])
\[ \Rightarrow r = \dfrac{4}{4} = 1m\]
Hence the radius of the circle is 1m.
Thus, option (a) is the right answer for the given question.
Note In a centripetal acceleration of the body, the acceleration which is directed along the radius of the path is called radial acceleration. In this case, the tangential velocity is said to be normal to the acceleration.
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