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# Rotation Problem watch

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1. 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?
2. 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.
3. How to get a formula for the tangential (linear) acceleration of the object in terms of r
and α ?

Can you show me a guided solution?

Thanks a lot.
4. The linear acceleration is r alpha.

so 2 r alpha = 2 r alpha theta
5. 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]
6. Can show me how to get linear/tangential acceleration = αr [r times the angular acceleration] ?
7. Differentiate s = theta x r (twice. )

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