Yahoo Answers: Answers and Comments for Centripetal Acceleration and Tangential Acceleration? [Physics]
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enAU
Thu, 16 Jun 2011 19:20:40 +0000
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Yahoo Answers: Answers and Comments for Centripetal Acceleration and Tangential Acceleration? [Physics]
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From Steve: ac = r*w² = 2r*α*Θ
at = r*α
For ac = 2at...
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https://au.answers.yahoo.com/question/index?qid=20110616192040AA10jAX
Thu, 16 Jun 2011 19:31:08 +0000
ac = r*w² = 2r*α*Θ
at = r*α
For ac = 2at,
2rαΘ = 2r*α
Θ = 1.0 rad
Interesting problem!

From electron1: An electric drill starts from rest and rotates...
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https://au.answers.yahoo.com/question/index?qid=20110616192040AA10jAX
Thu, 16 Jun 2011 20:02:33 +0000
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

From u.n. o: Let w = angular speed, rad/sec
Then, at some ...
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https://au.answers.yahoo.com/question/index?qid=20110616192040AA10jAX
Thu, 16 Jun 2011 20:05:17 +0000
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