Question

An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at x 2 m,its velocity is (4 m/s) .Give the object’s (a) velocity and (b) acceleration at y 2 m.

Answer 1

Answer:

(a) - 4 î m/s

(b) - 8 ĵ m/s²

Explanation:

Since, the center of the circular motion is at origin, and the particle crosses x=2 while doing circular motion, so, the radius must be 2 metre. Here, the speed of the object is 4 m/s.

Since, you haven't mentioned the direction of the object, or in which fashion (either clockwise or anticlockwise) the object is rotating. Let your object be going upwards at this point i.e., anticlockwise fashion.

So, it's velocity at x = 2 m will be v = + 4 ĵ m/s

Now, when the object will reach y = 2 m,

It's velocity will remain same, but the direction will be towards -ve x-axis, since I've let the object rotate in anti-clockwise fashion.

So, it's velocity at y = 2 m will be v = - 4 î m/s

Now, since the object is moving with constant speed, it's performing Uniform Circular Motion. So it's centripetal acceleration is given as

$a_{c} = \frac{v^{2} }{r}$

$a_{c} = \frac{4~*~4}{2}$

$a_{c} =$ 8 m/s²

Since, the centripetal acceleration is directed towards center, it's direction at point y = 2 m will be along -ve y-axis.

Hence, $a_{c} =$  - 8 ĵ m/s²