A particle moves in the x-y plane according to rule $\mathrm{x}=\mathrm{asin\omega t}$ and $\mathrm{y}=\mathrm{acos\omega t}$. The particle follows:

The magnitude of a vector $\overrightarrow{\mathrm{A}}$ is constant but it is changing its direction continuously. The angle between $\overrightarrow{\mathrm{A}}$ and $\frac{\mathrm{d}\overrightarrow{\mathrm{A}}}{\mathrm{dt}}$ is:

1. 180°

2. 120°

3. 90°

4. 0°

What is the value of linear velocity if \(\overset{\rightarrow}{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}\) and \(\overset{\rightarrow}{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}\)  :

The angle turned by a body undergoing circular motion depends on the time as given by the equation, $\theta ={\theta }_0+{\theta }_{1}t+{\theta }_2{t}^2$. It can be deduced that the angular acceleration of the body is? 

1. θ₁

2. θ₂

3. 2θ₁

4. 2θ₂

A particle moves in a circle of radius \(5\) cm with constant speed and time period \(0.2\pi\) s. The acceleration of the particle is:

If the equation for the displacement of a particle moving on a circular path is given by \(\theta = 2t^3 + 0.5\) where θ is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is:

1. 8 rad/sec

2. 12 rad/sec

3. 24 rad/sec

4. 36 rad/sec

A car moves on a circular path such that its speed is given by v = Kt, where K =constant and t is time. Also given: radius of the circular path is r. The net acceleration of the car at time t will be:

1. $\sqrt{{\mathrm{K}}^2 + {\left(\frac{{\mathrm{K}}^2{\mathrm{t}}^2}{\mathrm{r}}\right)}^2}$

2. 2K

3. K

4. $\sqrt{{\mathrm{K}}^2 + {\mathrm{K}}^2{\mathrm{t}}^2}$

Two particles A and B are moving in a uniform circular motion in concentric circles of radii \(r_A\) and \(r_B\) with speeds \(v_A\) and \(v_B\) respectively. Their time periods of rotation are the same. The ratio of the angular speed of \(A\)  to that of \(B\) will be:

A stone tied to the end of a 1 m long string is whirled in a horizontal circle at a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

The position vector of a particle is $\overrightarrow{\mathrm{r}}$ $=$ $\left(\mathrm{a}\right)$ $\sin$ $\mathrm{\omega t}\hat{\mathrm{i}}$ $+$ $\left(\mathrm{a}\right)$ $\cos$ $\mathrm{\omega t}\hat{\mathrm{j}}$. The velocity of the particle is: