Question

An object moves at a constant speed along a circular path in a horizontal $X-Y$ plane, with the centre at the origin. When the object is at $x=-2m$ , its velocity is $-4\hat{j}m{s}^{-1}$ What is the object's acceleration when it is $y=2m$ ?

• A. $-8\hat{j}m{s}^{-2}$
• B. $-8\hat{i}m{s}^{-2}$
• C. $-4\hat{j}m{s}^{-2}$
• D. $4\hat{i}m{s}^{-2}$

Answer 1

Answer: (A)

Here, radius of circular path, $r=2m$
Speed of the object, $v=4m{s}^{-1}$

The magnitude of centripetal acceleration
$a_c=\frac{v^2}{r}=\frac{4^2}{2}=8m{s}^{-2}$
This acceleration is directed towards the centre.
Therefore, when the object is at $y=2m,$
its position is $P$ (see figure).
Its acceleration is $-8\hat{j}m{s}^{-2}$.