A ring rotates about the z axis as shown in the figure. The plane of rotation is x-y. At a certain instant, the acceleration of a particle P is (6\hat{i}−8\hat{j})ms−2 . Find the angular acceleration of the ring and the angular velocity at that instant. Radius of the ring is 2m

 Tangential acceleration refers to the force acting along the tangent. This is commonly referred to as at. The magnitude is given by at=R, where R is the circular path’s radius.

The acceleration operating radially inwards towards the centre of the circular path described by the body is known as normal or centripetal acceleration. An is commonly used to represent this. The magnitude is provided by an = Rω2, where R is the circular path’s radius. The tangential acceleration will be equal to the x component of the specified acceleration at point P, whereas the normal acceleration will be equal to the y component.

Hence, we can say that at=Rα=6rads−2

Given that the radius of the circular path is 2m, we can find out the value of angular acceleration.

Substituting the values, we get,

6 = 2α

⇒α = 3rads−2

Also, we can say that an=Rω2 = 8rads−2

Given that the radius of the circular path is 2m, we can find out the value of angular velocity.

Substituting the values, we get,

8 = 2ω2

⇒ ω2 = 4

This gives us ω=2rads−1