A particle is constrained to move in a circle with a $10$ meter radius. At one instant, the particles’ speed is $10m/s$and is increasing at a rate of$10m/{s^2}$. The angle between the particle’s velocity and acceleration vectors is:
Answer
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Hint: We know that the rate at which the velocity of the particle increases is known as the acceleration of the particle. Just like velocity being a vector quantity, acceleration is also a vector quantity having two components. We need to analyse the components to arrive at a result.
Complete Step-By-Step Solution:
We know from the given question, that the particle moves in the circle. While moving, the particle encounters two types of acceleration, one is known as the centripetal acceleration and the other one is termed as tangential acceleration. These are basically the two components of the acceleration vector.
The particle here moves in the circular path, it experiences both centripetal and linear acceleration.
The tangential acceleration of the particle acts at an angle of ${90^o}$ with the diameter. The direction of this acceleration is tangent to the circular path.
The magnitude of tangential acceleration ${a_t}$ is $ = 10m/{s^2}$
The magnitude of centripetal acceleration is given by $ = \dfrac{{{v^2}}}{r}$
Where, $v$ is the velocity of the particle
$r$ is the radius of the circular path.
Thus, ${a_c} = \dfrac{{(10 \times 10)}}{{10}} = 10m/{s^2}$
Thus both ${a_c}$ and ${a_t}$ are both of the same magnitude, thus making equal angle with the resultant.
Since ${a_t}$ is tangent to the circular path, ${a_t}$ makes an angle ${45^o}$with the resultant. Thus, ${a_c}$ also makes angle ${45^o}$ with the resultant.
Direction of tangential acceleration and velocity is the same, so we can say, the angle made with the velocity is ${45^o}$.
Note:
We know, tangential acceleration is defined as the change in tangential velocity of a point around a radius with change in time as a body moves in a circular path.
Centripetal acceleration is defined as the property of a body moving in a circular path, this acceleration moves radially directed towards the centre.
Complete Step-By-Step Solution:
We know from the given question, that the particle moves in the circle. While moving, the particle encounters two types of acceleration, one is known as the centripetal acceleration and the other one is termed as tangential acceleration. These are basically the two components of the acceleration vector.
The particle here moves in the circular path, it experiences both centripetal and linear acceleration.
The tangential acceleration of the particle acts at an angle of ${90^o}$ with the diameter. The direction of this acceleration is tangent to the circular path.
The magnitude of tangential acceleration ${a_t}$ is $ = 10m/{s^2}$
The magnitude of centripetal acceleration is given by $ = \dfrac{{{v^2}}}{r}$
Where, $v$ is the velocity of the particle
$r$ is the radius of the circular path.
Thus, ${a_c} = \dfrac{{(10 \times 10)}}{{10}} = 10m/{s^2}$
Thus both ${a_c}$ and ${a_t}$ are both of the same magnitude, thus making equal angle with the resultant.
Since ${a_t}$ is tangent to the circular path, ${a_t}$ makes an angle ${45^o}$with the resultant. Thus, ${a_c}$ also makes angle ${45^o}$ with the resultant.
Direction of tangential acceleration and velocity is the same, so we can say, the angle made with the velocity is ${45^o}$.
Note:
We know, tangential acceleration is defined as the change in tangential velocity of a point around a radius with change in time as a body moves in a circular path.
Centripetal acceleration is defined as the property of a body moving in a circular path, this acceleration moves radially directed towards the centre.
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