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Question: A wheel has angular acceleration of 3rad\({{s}^{-2}}\) and an initial angular speed of 2rad\({{s}^{-...

A wheel has angular acceleration of 3rads2{{s}^{-2}} and an initial angular speed of 2rads1{{s}^{-1}}. In a time of 2s it has rotated through an angle (in radians) of:
A. 6
B. 10
C. 12
D. 4

Explanation

Solution

Rotational motion is analogous to translation motion. Therefore, we can relate θ\theta , ω\omega and α\alpha as θ=ω0t+12αt2\theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}}. Use this formula to find the angle rotated by the wheel in 2 seconds.

Formula used:
θ=ω0t+12αt2\theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}}

Complete answer:
When a body is in rotational motion about a fixed axis called axis of rotation, the angle of rotation (θ\theta ) of the body continuously changes. The rate of change of angle of rotation of the body with respect to time is called angular velocity of the body. It is equal to the change of the angle in one unit time. It is denoted by ω\omega .
Then, when the angular velocity of the body is changing, we define angular acceleration of the body. Angular acceleration is defined as the rate of change of angular velocity of the body with respect to time. It is equal to the change of angular velocity of the body in one unit of time. It is denoted by α\alpha .
Rotational mechanics is analogous to translational mechanics.
The angle θ\theta is analogous to displacement s.
The angular velocity ω\omega is analogous to velocity v.
The acceleration α\alpha is analogous to acceleration a.
Hence, the relation between θ\theta , ω\omega , α\alpha is the same as the relation between s, v, a.
When a body is in pure translational motion, we know s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}, where u is the initial velocity of the body and t is the given time.
Hence, if the body is in pure rotational motion θ=ω0t+12αt2\theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}} …… (i),
where ω0{{\omega }_{0}} is the initial angular velocity of the body.
In the given case, α=3s2\alpha =3{{s}^{-2}}, ω0=2s1{{\omega }_{0}}=2{{s}^{-1}}, and t=2s.
Substitute the values in equation (i).
θ=2(2)+123(2)2=4+6=10rad\Rightarrow \theta =2(2)+\dfrac{1}{2}3{{(2)}^{2}}=4+6=10rad
Hence, the correct option is B.

Note: Note that like displacement, velocity and acceleration are vectors, angle of rotation, angular velocity and angular acceleration are also vectors.
The direction of these vectors are perpendicular to the position vector of the body (vector joining the position of the body and the axis rotation).