J asked in Science & MathematicsPhysics · 9 years ago

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)

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  • 9 years ago
    Best 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

  • u.n. o
    Lv 5
    9 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

  • Steve
    Lv 7
    9 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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