# A peg moving on a ring [closed]

A thin but rigid semicircular wire frame of radius r is hinged at O and can rotate in its own vertical plane. A smooth peg P starts from O and moves horizontally with constant speed v0, lifting the frame upward as shown in figure. Find the angular velocity of the frame when its diameter makes an angle of 60° with the vertical :

This was an interesting question I came across

My work: Here, for the ring to be pulling upward completely it would have to be rotated by an angle pi/6. wt=pi/6 The horizontal displacement when the ring is rotated would be vt=(2-√3)r Thus w=pi/6(v(2-√3)r)

The options were a)v/r b)v/2r c)2v/r d) vr

This is one of those questions where I dont get much ideas on how to solve. What method should i be using to solve this. I would really appreciate it if somebody helps me solve this question and add new techniques of problem solving techniques to my arsenal. That would help me look at the world in a different way.

## closed as off-topic by tpg2114♦May 18 at 12:08

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• Is there any gravity? – Archisman Panigrahi May 18 at 6:05
• Yes! i also tried balancing out forces by drawing Free body diagrams. I just cant find w. – Abi Nand May 18 at 6:07
• A quick method to solve this problem would be to use the property of rigid body that angular velocity is same as observed from different points on that body. This would solve the problem orally I.e. V/r. – Unique May 19 at 18:38

Let $$x$$ denote the distance between $$P$$ and $$O$$.

$$\frac d{dt}x=v_0$$

Let $$\theta$$ denote the angle the frame's diameter makes with the vertical. In the picture $$\theta=60^\circ$$ equals $$\frac\pi3$$ radians. Let $$\theta'$$ denote the angular velocity of the frame.

$$\frac d{dt}\theta=\theta'$$

It takes two steps to find $$\theta'$$ in terms of $$v_0$$. The first step is to find $$\theta$$ in terms of $$x$$. Then we take the derivative of $$x$$ and $$\theta$$ with respect to time $$t$$ and we have our answer. In other words, the way to solve this problem is to do geometry and then do calculus.

A straight line segment whose endpoints both lie on a circle is called a chord. The length of a chord is $$r\mathrm{crd}(a)=r2\sin\frac a2$$ where $$a$$ is the angle of the chord. We must now find a relationship between $$a$$ and $$\theta$$.

Let $$o$$ denote the origin of the circular wire. The angle $$a$$ is defined to be the angle $$OoP$$. A vertical line bisects $$a$$. All vertical lines are parallel to each other. Thus $$\theta$$ is alternate interior angles to $$\frac a2$$ which implies $$\theta=\frac a2$$.

$$2\theta=a$$

Plug $$2\theta=a$$ into the chord equation and then take the derivative with respect to time and we have the answer.

$$x=r2\sin\frac a2=r2\sin\theta$$ $$\frac d{dt}x=\frac d{dt}r2\sin\theta$$ $$v_0=\theta'r2\cos\theta=\theta'r2\cos\frac\pi3=\theta'r2\frac12=\theta'r$$ $$\theta'=\frac{v_0}r$$