# Motion down an inclined plane with leg equal to the diameter of a circle

This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach".

An object slides down an inclined plan $$OP'$$ (Fig. 3-9) starting from rest at $$0$$. Show that the point $$Q$$ reached in the time $$t_1$$ required to fall straight down to $$P$$ lies on a circle with $$OP$$ as diameter.

The text gives a proof by contradiction for this problem:

From the formula $$s = 16t^2$$ we find that the time to fall the distance $$OP$$ is $$t_1 = \sqrt{OP}/4$$. For the motion along $$OP'$$ we use (35), that is, $$s = 16t^2 \sin{A}$$. The distance $$OQ$$ that the object slides in time $$t_1$$ is $$OQ = 16(OP/16) \sin{A}$$. Then $$\sin{A} = OQ / OP$$. Suppose $$Q$$ is not on the circle but R on $$OP'$$ is. Then $$\angle OPR$$ is $$\angle A$$ by the use of right triangles. Then $$\sin A = OR/OP$$. But $$\sin{A} = OQ/OP$$. Hence $$Q = R$$ and $$Q$$ lies on the circle.

This proof is confusing to me because I do not know the logical form of the statement we are trying to prove. What if $$Q$$ does not intersect with the circle at all? What is the justification for the bold portions of the proof?

Is it possible to directly show that $$Q$$ lies on the circle?

My approach is to let $$C$$ be the center of the circle. If we can show that the length of $$CO$$ is equal to the length of $$CQ$$, then $$Q$$ will be on the circle. Although, I keep getting stuck trying to show that $$CQ = \frac{OP}{2}$$.

• Did you forget that $OP = 2CO$?
– YNK
Jul 8, 2020 at 15:22
• @YNK I'm still a bit confused on how to use that. Any more hints? Jul 8, 2020 at 16:05
• I am sorry. My comment is wrong. Forget about it. Give me an hour or so. I will see what I can do.
– YNK
Jul 8, 2020 at 16:26

## 5 Answers

Suppose $$Q$$ is not on the circle but R on $$OP'$$ is.

This part supposes $$Q\not=R$$ where $$R$$ is the intersection point of the circle with $$OP'$$ where $$R\not=O$$. (note that $$Q$$ is on $$OP'$$.)

Then $$\angle OPR$$ is $$\angle A$$ by the use of right triangles.

Since $$\triangle{OPP'}$$ and $$\triangle{PRP'}$$ are right triangles, we get $$\angle{OPR}=\angle{OPP'}-\angle{RPP'}=90^\circ-\angle{RPP'}=(180^\circ-\angle{PRP'})-\angle{RPP'}=\angle A$$

The last step of the proof is as follows :

It follows from $$\sin A=\frac{OQ}{OP}$$ and $$\sin A=\frac{OR}{OP}$$ that $$\frac{OQ}{OP}=\frac{OR}{OP}\implies OQ=OR\implies Q=R$$ which contradicts the supposition that $$Q\not=R$$. So, we see that $$Q=R$$, and that $$Q$$ lies on the circle.

By the way, I think we can prove that without using a proof by contradiction as follows :

(After getting $$\sin A=\frac{OQ}{OP}$$) Let us define $$R$$ as the intersection point of the circle with $$OP'$$ where $$R\not=O$$. Then, we get $$\sin A=\frac{OR}{OP}$$. It follows that $$\frac{OQ}{OP}=\frac{OR}{OP}\implies OQ=OR\implies Q=R$$. So, $$Q$$ is on the circle.

Is it possible to directly show that $$Q$$ lies on the circle? My approach is to let $$C$$ be the center of the circle. If we can show that the length of $$CO$$ is equal to the length of $$CQ$$, then $$Q$$ will be on the circle. Although, I keep getting stuck trying to show that $$CQ = \frac{OP}{2}$$.

Applying the law of cosines to $$\triangle{OQC}$$, we get \begin{align}CQ&=\sqrt{OQ^2+OC^2-2OQ\cdot OC\cos\angle{QOC}} \\\\&=\sqrt{(OP\sin A)^2+\bigg(\frac{OP}{2}\bigg)^2-2\cdot OP\sin A\cdot\frac{OP}{2}\cos(90^\circ -A)} \\\\&=\sqrt{OP^2\sin^2A+\bigg(\frac{OP}{2}\bigg)^2- OP^2\sin^2A} \\\\&=\frac{OP}{2}\end{align}

Assume that $$Q$$ lies on the circle and then join $$PQ$$. We have $$\angle PQO=90°$$ and hence $$\angle QPO=A$$. Now in $$\triangle OPQ$$, $$\sin A=\frac{OQ}{OP}$$ This result is same as you derived without considering right angle at $$Q$$ (which is correct). So, our assumptions was correct and $$Q$$ lies on the circle.

• If we assume that $Q$ lies on the circle, how does it follow that $\angle PQO = 90$ degrees? Jul 9, 2020 at 2:13
• Angle subtended at circumference is half of angle subtended at center of the circle. If $C$ is center of the circle then $\angle OCP=180°\implies \angle PQO=90°$. Jul 9, 2020 at 2:50

With a few Physics.

The vertical acceleration $$(0,-g)$$ projected over $$OP'$$ is $$-\sin A(\cos A,\sin A)g$$ Now the movement along $$OQ$$ is given by

$$(x,y) = (0,|OP|)-\frac 12m \sin A(\cos A,\sin A)gt^2$$

but $$t$$ is the time needed to reach $$P$$ in the vertical fall so

$$|OP| = \frac 12 m g t^2$$

so eliminating $$t$$ we obtain

$$(x,y) = (0,|OP|)-\sin A(\cos A,\sin A)|OP|$$

and we have

$$\cases{ x = -\sin A\cos A|OP|\\ y = (1-\sin^2 A) |OP| }$$

and $$(x(A), y(A))$$ describes a circle as expected.

The gravitational acceleration is $$g$$ from O straight down to P, while it is $$g\cos\theta$$ from O to Q along the ramp. Use the time-distance formula $$d = \frac12 a t^2$$ for an accelerating object to establish the equation of same time for the two cases,

$$t^2 = \frac{2OP}g = \frac{2OQ}{g\cos\theta}$$

Simplify to get $$\cos\theta = \frac{OQ}{OP}$$, which indicates that $$\triangle$$OQP is a right triangle. Hence, Q lies on the circumcircle of $$\triangle$$OQP with OP as the diameter.

For both paths.

Time = $$\frac 12 \frac {\text {distance}}{\text {acceleration}}$$

On the trip from $$O$$ to $$Q,$$ the acceleration is $$g\sin A$$

The distance covered is $$\overline {OP} \sin A$$

The alternative trip from $$O$$ to $$P$$ the accelartion is $$g$$ and the distance is $$\overline {OP}$$