Uniform circular motion  (Page 5/5)

In order to determine centripetal acceleration, we need to find the speed of the particle. We can determine the speed of the particle, using expression of time period :

$$\begin{array}{l}T=\frac{2\pi r}{v}\end{array}$$

$$\begin{array}{l}\Rightarrow v=\frac{2\pi r}{T}\end{array}$$

The acceleration of the particle is :

$$\begin{array}{lll}a=\frac{v^2}{r}& =\frac{{\left(2\pi r\right)}^2}{T^{2}r}& \frac{4{\pi }^{2}r}{T^2}\end{array}$$

Hence, option (a) is correct.

Three quantities, namely, radius of the circular path, magnitude of acceleration and speed are constant in uniform circular motion.

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

The options (a), (c) and (d) are wrong.

Hence, option (b) is correct.

The velocity is a vector quantity. We need to know the components of the velocity when particle makes the angle 135° with x - axis.

Uniform circular motion

$$\begin{array}{l}{v}_x=-v\sin \theta =-10\sin {135}^0=-10x\left(\frac{1}{\sqrt{2}}\right)=-5\sqrt{2}\\ v_y=v\cos \theta =10\cos {135}^0=10x\left(\frac{-1}{\sqrt{2}}\right)=-5\sqrt{2}\end{array}$$

The component $v_x$ is in the reference x - direction, whereas component $v_y$ is in the negative reference y - direction.

$$\begin{array}{l}\mathbf{v}=v_x\mathbf{i}+v_y\mathbf{j}\\ \Rightarrow \mathbf{v}=-(5\sqrt{2}\mathbf{i}+5\sqrt{2}\mathbf{j})m/s\end{array}$$

Hence, option (a) is correct.

The curvatures of path at four points are indicated with our circles drawn so that they align exactly with the curved path at these points. The radii of these circles at points A and B are smaller than that at points C and D.

Uniform circular motion

Now, the centripetal acceleration is given by :

$$\begin{array}{l}a=\frac{v^2}{r}\end{array}$$

The smaller the radius, greater is centripetal acceleration. Thus,

$$\begin{array}{l}{a}_A>a_D\end{array}$$

and

$$\begin{array}{l}{a}_B>a_D\end{array}$$

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

The component of velocity along x - direction is “-v sinθ” and is not a constant. The component of acceleration along x - direction is “ $a_x=-\frac{v^2}{r}\cos \theta$ ” and is not a constant. The external force in uniform circular motion is equal to centripetal force, which is directed towards the center. Thus, the velocity is perpendicular to external force.

On the other hand, the particle covers equal circular arc in equal time interval. Hence, the angle subtended by the arc at the center is also equal in equal time interval.

Hence, option (c) is correct.

From the figure, it is clear that the components of velocity in the fourth quadrant are both positive.

Uniform circular motion

Hence, option (d) is correct.

Here, the inputs are given in terms of coordinates. Thus, we shall use the expression of velocity, which involves coordinates. Velocity of the particle in circular motion is given by :

Uniform circular motion

$$\begin{array}{l}\mathbf{v}=-\frac{vy}{r}\mathbf{i}+\frac{vx}{r}\mathbf{j}\end{array}$$

Here, radius of the circle is :

$$\begin{array}{l}r=\sqrt{(3^2+4^2)}=5m/s\end{array}$$

$$\begin{array}{l}\Rightarrow \mathbf{v}=-\frac{1x(-4)}{5}\mathbf{i}+\frac{1x3}{5}\mathbf{j}\end{array}$$

$$\begin{array}{l}\Rightarrow \mathbf{v}=\frac{1}{5}(4\mathbf{i}+3\mathbf{j})\end{array}$$

Hence, option (a) is correct.