Rotational Motion Question 9
Question 9 - 29 January - Shift 2
An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at $x=+2 m$, its velocity is $-4 \hat{j} m / s$. The object’s velocity (v) and acceleration (a) at $x=-2 m$ will be
(1) $v=4 \hat{i} m / s, a=8 \hat{j} m / s^{2}$
(2) $v=4 \hat{j} m / s, a=8 \hat{i} m / s^{2}$
(3) $v=-4 \hat{j} m / s, a=8 \hat{i} m / s^{2}$
(4) $v=-4 \hat{i} m / s, a=-8 \hat{j} m / s^{2}$
Show Answer
Answer: (2)
Solution:
Formula: Centripetal acceleration
$a_c=\frac{V^{2}}{r}=\frac{4^{2}}{2}=\frac{16}{2}=8 m / s^{2}$
$\overrightarrow{{}V}=4 \hat{j}$
$\overrightarrow{{}a_c}=8 \hat{i}$
