An electric drill
startsfrom rest and rotates with a constant angular acceleration.
afterthe drill has rotated through a certain angle, the magnitude
of thecentripetal acceleration of a point on the drill is the
magnitudeof the tangential acceleration. what is theangle?

Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.

The answer is :  Δ θ = 1 rad

Explanation:

Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:

Ac= centripetal acceleration            At= tangential acceleration

Ac = V² / r                                                At = r α  

Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration

V = r ω

Ac = (r ω)² / r = r² ω² / r

Ac = r ω²

now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.

Ac = 2 At

That is,  

r ω² = 2 r α

it is equivalent to;

ω²  = 2 α

now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,  

ω² – ω°² = 2 α Δ θ

such that ω° = 0

and ω² = 2 α

therefore;

2 α - 0 = 2 α Δ θ

2 α = 2 α Δ θ

So the angle will be :  Δ θ = 1 rad

the components of a vector are the parts of the vectors that are to each other.

b perpendicular