Answer:
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
