6.2 Circular motion and rotational kinematics  (Page 6/6)

Flywheel covers angular displacement of "2π" in one revolution. Therefore, the angular speed in "radian/second" is :

$$\omega =\frac{2\pi X300}{60}=10\pi \mathrm{rad}/s$$

Hence, option (b) is correct.

The time period of rotation is :

$$T=\frac{2\pi r}{v}=\frac{2\pi }{\omega }$$ $$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2s$$

Note : The input of radius is superfluous here.

Hence, option (b) is correct.

The time period of rotation is :

$$T=\frac{2\pi r}{v}=\frac{2\pi }{\omega }$$

Rearranging, the angular speed is :

$$\Rightarrow \omega =\frac{2\pi }{T}=\frac{2\pi }{1}=2\pi$$

Converting to in the unit of revolution per second, the angular speed is :

$$\Rightarrow \omega =2\pi \text{rad /s}=\frac{2\pi }{2\pi }=1\text{revolution/s}$$

Hence, option (a) is correct.

The linear velocity is given by the following vector cross product,

$$\mathbf{v}=\mathbf{\omega }X\mathbf{r}$$

We know that a change in the order of operands in cross product reverses the resulting vector. Hence,

$$\mathbf{v}=-\mathbf{r}X\mathbf{\omega }$$

Further, it is given in the question that positive axial direction has the unit vector "n". But, the plane of motion is perpendicular to axis of rotation. Hence, velocity is not aligned in the direction of unit vector n .

Hence, options (a) and (b) are correct.

The direction of rotation is determined by the sign of angular velocity. In turn, the sign of angular velocity is determined by the sign of the slope on angular displacement - time plot. The sign of slope is negative for line OA, positive for line AC and zero for line CD.

The positive angular velocity indicates anti-clockwise rotation and negative angular velocity indicates clockwise rotation. The disk is stationary when angular velocity is zero.

Hence, options (b) and (c) are correct.

In order to find angular displacement, we need to find initial and final angular positions. From the geometry,

Angular displacement

$$\tan {\theta }_1=\frac{\sqrt{3}}{1}=\sqrt{3}=\tan {60}^0$$ $$\Rightarrow {\theta }_1={60}^0$$

Similarly,

$$\tan {\theta }_2=-\frac{\sqrt{3}}{1}=-\sqrt{3}=\tan {120}^0$$ $$\Rightarrow {\theta }_2={120}^0$$

Thus, angular displacement is :

$$\Rightarrow \Delta \theta ={\theta }_2-{\theta }_1={120}^0-{60}^0={60}^0$$

Hence, option (c) is correct.

The direction of linear velocity is normal to radial direction. The direction of centripetal acceleration is radial. The direction of angular velocity is axial. See the figure.

Angular motion

Clearly, the three directions are mutually perpendicular to each other.

Hence, options (c) and (d) are correct.

The centripetal acceleration is given by :

$$a_R=\frac{v^2}{r}$$

The linear speed "v" is related to angular speed by the relation,

$$\Rightarrow v=\omega r$$

Substituting this in the expression of centripetal acceleration, we have :

$$\Rightarrow a_R=\frac{{\omega }^2{r}^2}{r}={\omega }^{2}r$$

We need to evaluate fourth cross product expression to know whether its modulus is equal to the magnitude of centripetal acceleration or not. Let the cross product be equal to a vector " A ". The magnitude of vector " A " is :

$$A=\left|\mathbf{\omega }X\mathbf{v}\right|=\omega v\sin \theta$$

For circular motion, the angle between linear and angular velocity is 90°. Hence,

$$\Rightarrow A=\left|\mathbf{\omega }X\mathbf{v}\right|=\omega v\sin {90}^0=\omega v$$

Substituting for " ω", we have :

$$\Rightarrow A=\omega v=\frac{v^2}{r}$$

The modulus of the expression, therefore, is equal to the magnitude of centripetal acceleration.

Note : The vector cross product "| ω X v |” is the vector expression of centripetal force.

Hence, options (a) and (d) are correct.