The velocity and acceleration vectors of a pa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In circular motion,the acceleration vector $\overrightarrow{a}$ can be resolved into two components: the centripetal acceleration $\overrightarrow

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) In circular motion,the acceleration vector $\overrightarrow{a}$ can be resolved into two components: the centripetal acceleration $\overrightarrow{a}_c$ (perpendicular to velocity) and the tangential acceleration $\overrightarrow{a}_t$ (parallel to velocity).Given $\overrightarrow{v} = 2 \hat{i} 	ext{ m/s}$ and $\overrightarrow{a} = 2 \hat{i} + 4 \hat{j} 	ext{ m/s}^2$.The component of acceleration perpendicular to the velocity is the centripetal acceleration,$a_c$.Since $\overrightarrow{v}$ is along the $\hat{i}$ direction,the component of $\overrightarrow{a}$ perpendicular to $\overrightarrow{v}$ is the $\hat{j}$ component,which is $a_c = 4 	ext{ m/s}^2$.The formula for centripetal acceleration is $a_c = \frac{v^2}{R}$,where $R$ is the radius of the circle.Rearranging for $R$,we get $R = \frac{v^2}{a_c}$.Substituting the values: $R = \frac{(2)^2}{4} = \frac{4}{4} = 1 	ext{ m}$.

The velocity and acceleration vectors of a particle undergoing circular motion are $\overrightarrow{v} = 2 \hat{i} \text{ m/s}$ and $\overrightarrow{a} = 2 \hat{i} + 4 \hat{j} \text{ m/s}^2$ respectively at an instant of time. The radius of the circle is $........ \text{ m}$.

Similar Questions

$A$ particle of mass $5 \ kg$ is moving on a circle of radius $1 \ m$ with an angular velocity of $2 \ rad \ s^{-1}$. The centripetal force is (in $N$)

$A$ stone is tied to one end of a string $50\, cm$ long and is whirled in a horizontal circle with a constant speed. If the stone makes $10$ revolutions in $20\, s$,what is the magnitude of the acceleration of the stone in $cm/s^2$?

$A$ particle moves along a circle of radius $3 \, m$ with its displacement given by $S = \frac{t^2}{2} + \frac{t^3}{3}$. The total acceleration at $t = 2 \, s$ is ....... $m/s^2$.

$A$ particle is in uniform circular motion. The equation of its trajectory is given by $(x-2)^2+y^2=25$,where $x$ and $y$ are in meters. The speed of the particle is $2 \text{ m/s}$. When the particle attains the lowest $y$ coordinate,the acceleration of the particle is (in $\text{m/s}^2$):

For a particle in circular motion,$a_r = 3 \ m/s^2$ and $a_T = 4 \ m/s^2$. If $\theta$ is the angle between the resultant acceleration $a$ and the radial acceleration $a_r$,then .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module