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Rotation Problem watch

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    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?
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    The centripetal acceleration of an object moving in a circle of radius r with velocity v is
    v²/r
    v=rω where ω is the angular velocity of the drill.
    So centripetal acceleration = rω²

    If the drill is accelerating with angular acceleration α then using ω2²=ω1² + 2αθ
    [This is the "v²=u² + 2as" equation for rotational acceleration.]
    gives ω²=2αθ where θ is the angle turned through. (Initial ω is zero)
    This gives
    centripetal acceleration = 2rαθ
    Now get a formula for the tangential (linear) acceleration of the object in terms of r and α, and find the value of θ such that the centripetal is twice the tangential.
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    How to get a formula for the tangential (linear) acceleration of the object in terms of r
    and α ?

    The answer given is 1.00 rad.

    Can you show me a guided solution?

    Thanks a lot.
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    The linear acceleration is r alpha.

    so 2 r alpha = 2 r alpha theta
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    The next step is very straightforward.
    The linear/tangential speed v at radius r is ωr [r times angular speed]
    The linear/tangential acceleration = αr [r times the angular acceleration]
    So from my 1st post you have
    centripetal acceleration = 2rαθ
    and from this post you have
    linear acceleration = αr
    You really need to do the last step yourself.
    Put the centripetal = 2 times tangential and see what the value of theta is. [Hint: all the other terms cancel out]
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    Can show me how to get linear/tangential acceleration = αr [r times the angular acceleration] ?
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    Differentiate s = theta x r (twice. )
 
 
 
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