# Centripetal Acceleration and Tangential Acceleration?

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. What is the angle? (in radians)

### 3 Answers

- electron1Lv 79 years agoFavourite answer
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. What is the angle? (in radians)

Centripetal acceleration is the acceleration caused by centripetal force.

Centripetal force = mass * (tangential velocity)^2 ÷ radius

Centripetal acceleration = (tangential velocity)^2 ÷ radius

Tangential acceleration = linear acceleration of a point on the circumference of the circle.

Linear acceleration = angular acceleration * radius

Tangential acceleration = angular acceleration * radius

Let’s say the radius = r meters and the angular acceleration = α radians/second^2.

The tangential acceleration of a point on the circumference of the circle = (α * r) m/s^2

The centripetal acceleration = 2 * (α * r) m/s^2

Centripetal acceleration = (tangential velocity)^2 ÷ radius = v^2 ÷ r

2 * (α * r) = v^2 ÷ r

v^2 = 2 * α * r^2

Angular acceleration = tangential acceleration ÷ radius

α = a ÷ r

v^2 = 2 * (a ÷ r) * r^2

v^2 = 2 * a * r

v = (2 * a * r)^0.5

As the drill accelerates, the velocity increases. You need to determine the distance the point on the circumference of the circle has moved when the velocity of the point = (2 * a * r)^0.5

vf^2 – vi^2 = 2 * a * d

vi = 0

vf = (2 * a * r)^0.5

vf^2 = (2 * a * r)

(2 * a * r) – 0 = 2 * a * d

(2 * a * r) = 2 * a * d

d = r

When the point on the circumference of the circle has moved a distance equal to the radius, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration.

The angle = 1 radian

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- u.n. oLv 59 years ago
Let w = angular speed, rad/sec

Then, at some time t = T, w^2*r = r*(dw/dt), r = radius of bit

So, w^2 = 2dw/dt

Let a = angular accel. = constant, rad/s^2

w = aT

Then, (aT)^2 = 2*(dw/dt) = 2a

a = 2/T^2

Let total angle traveled = theta = (a*T^2)/2

Then theta = (a*T^2)/2 = (T^2/2)*(2/T^2) = 1 radian Answer

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- SteveLv 79 years ago
ac = r*w² = 2r*α*Θ

at = r*α

For ac = 2at,

2rαΘ = 2r*α

Θ = 1.0 rad

Interesting problem!

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