Two frames of references S and S′ have a common origin O and S′ rotates with constant angular velocity ω with respect to S. A square hoop ABCD is made of fine smooth wire and has side length 2a. The hoop is horizontal and rotating with constant angular speed ω about a vertical axis through A. A small bead which can slide on the wire is initially at rest at the midpoint of the side BC. Choose axes relative to the hoop and let y be the distance of the bead from the vertex B on the side BC. Write down the position vector of the bead in your rotating frame. Show that y¨−ω2y=0 using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex C.

Megan Herman

Megan Herman

Answered question

2022-09-27

Two frames of references S and S have a common origin O and S rotates with constant angular velocity ω with respect to S.
A square hoop A B C D is made of fine smooth wire and has side length 2 a. The hoop is horizontal and rotating with constant angular speed ω about a vertical axis through A. A small bead which can slide on the wire is initially at rest at the midpoint of the side B C. Choose axes relative to the hoop and let y be the distance of the bead from the vertex B on the side B C. Write down the position vector of the bead in your rotating frame. Show that y ¨ ω 2 y = 0 using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex C.
I showed that d 2 r d t 2 = ( d 2 r d t 2 ) + 2 ω × ( d r d t ) + ω × ( ω × r ) where ′ indicates that it's done in rotating frame. r is position vector of a point P measured from the origin.
So,
r = r cos θ i + y j
r = ( r ˙ c o s θ r θ ˙ sin θ ) i + y ˙ j
r = ( r ¨ c o s θ r ˙ θ ˙ s i n θ r ˙ θ ˙ sin θ r θ ¨ sin θ r θ ˙ 2 c o s θ ) i + y ¨ j
ω × r = ω y ˙ i + ( ω r ˙ cos θ ω r θ ˙ sin θ ) j
ω × ( ω × r ) = ω 2 r cos θ i ω 2 y j

Answer & Explanation

AKPerqk

AKPerqk

Beginner2022-09-28Added 9 answers

When differentiating r = r cos θ i + y j , you have considered r to be constant, which is wrong. r is given by
r = l 2 + y 2
where l is the side-length of the square. So r will change with y, and you'll have to differentiate r too.
This is where the math gets pretty ugly.
To avoid that, what we can do is we can observe that in the rotation frame, the bead will experience an outward centrifugal force. This force will have a component along B C. That component can be written as
F B C = m ω 2 r sin θ
F B C = m ω 2 l 2 + y 2 y l 2 + y 2
Thus by dividing by m on both sides you get
y ¨ = ω 2 y
Note that this is the same as applying d 2 r d t 2 = ( d 2 r d t 2 ) + 2 ω × ( d r d t ) + ω × ( ω × r ). It is just that this approach is more problem specific.

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