Question
Read the passage given below and answer the following questions from (i) to (v). When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word
uniform refers to the speed which is uniform (constant) throughout the motion. Although the speed does not vary, the particle is accelerating because the velocity changes its direction at every point on the circular track. The figure shows a particle P which moves along a circular track of radius r with a uniform speedu.
  1. A circular motion:
  1. Is one-dimensional motion.
  2. Is two-dimensional motion.
  3. It is represented by combination of two variable vectors.
  4. Both (b) and (c)
  1. For a particle performing uniform circular motion, choose the incorrect statement from the following.
  1. Magnitude of particle velocity (speed) remains constant.
  2. Particle velocity remains directed perpendicular to radius vector.
  3. Direction of acceleration keeps changing as particle moves.
  4. Angular momentum is constant in magnitude but direction keeps changing.
  1. Two cars A and B move along a concentric circular path of radius $r_A$ and $r_B$ with velocities $v_A$ and $v_B$ maintaining constant distance, then $\frac{\text{v}_{\text{A}}}{\text{v}_\text{B}}$ is equal to:
  1. $\frac{\text{r}_{\text{B}}}{\text{r}_\text{A}}$
  2. $\frac{\text{r}_{\text{A}}}{\text{r}_\text{B}}$
  3. $\frac{\text{r}_{\text{A}}^2}{\text{r}_\text{B}^2}$
  4. $\frac{\text{r}_{\text{B}}^2}{\text{r}_\text{A}^2}$
  1. A car runs at a constant speed on a circular track of radius 100m, taking 62.8s for every circular lap. The average velocity and average speed for each circular lap, respectively is:
  1. $0,0$
  2. $0,10ms^{-1}$
  3. $10ms^{-1}, 10ms^{-1}$
  4. $10ms^{-1}, 0$
  1. A particle is revolving at 1200 rpm in acircle of radius 30cm. Then, its acceleration is:
  1. $1600ms^{-2}$
  2. $4740ms^{-2}$
  3. $2370ms^{-2}$
  4. $5055ms^{-2}$

Answer

  1. (d) Both (b) and (c)
Explanation:
Circular motion is an example of two-dimensional motion with radius vector as
$\text{r}=\text{a}\cos\omega\text{t}\hat{\text{i}}+\text{a}\sin\omega\text{t}\hat{\text{j}}$
Both the components $\text{r}=\text{a}\cos\omega\text{t}\hat{\text{i}}$ and $\text{a}\sin\omega\text{t}\hat{\text{j}}$ are perpendicular to each other.
  1. (c) Direction of acceleration keeps changing as particle moves.
Explanation:
If a particle is performing uniform circular motion, then its
  1. Speed will be constant throughout the motion.
  2. Velocity will be tangential in the direction of motion at a particular point.
  1. Acceleration, $\text{a}=\frac{\text{v}^2}{\text{r}}$ will always be towards centre of the circular path.
  2. Angular momentum (mvr) is constant in magnitude but direction keeps on changing.
  1. (b) $\frac{\text{r}_{\text{A}}}{\text{r}_\text{B}}$
Explanation:
Angular velocity $\omega$ is constant.
$\text{v}=\text{r}\omega$
$\therefore\text{v}\propto\text{r}$ or $\frac{\text{v}_\text{A}}{\text{v}_\text{B}}=\frac{\text{r}_{\text{A}}}{\text{r}_{\text{B}}}$
  1. (b) $0,10ms^{-1}$
Explanation:
On a circular path in completing one turn,the distance travelled is $2\pi\text{r},$ while displacement
is zero. Hence, average velocity
$=\frac{\text{displacement}}{\text{time interval}}=\frac{0}{\text{t}}=0$
Average speed $=\frac{\text{Distance}}{\text{Time interval}}$
$=\frac{2\pi\text{r}}{\text{t}}=\frac{2\times3.14\times100}{62.8}=10\text{ms}^{-1}$
  1. (b) $4740ms^{-2}$
Explanation:
Given, V = 1200 rpm $\frac{1200}{60}\text{rps}$
$\text{r}=30\text{cm}=\frac{30}{100}\text{m}$
Acceleration of the particle = Centripetal acceleration
$=\omega^2\text{r}=(2\pi\text{v})^2\text{r}$
$=\Big(2\times\frac{22}{7}\times\frac{1200}{60}\Big)^2\times\frac{30}{100}\approx4740\text{ms}^{-2}$

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