# Tutor profile: Elena T.

## Questions

### Subject: Physics (Waves and Optics)

An observer standing by the edge of a pond perceives it as 1 m deep near the shoreline. A refractive index of water is $$n=1.33$$. What is the true depth of the pond?

Let $$A$$ be a point on the bottom of the pond; $$B$$ be a point on the surface directly above $$A$$; and $$C$$ be a point on the surface at a small distance from $$B$$. Think of a light ray directed from $$A$$ to $$C$$. This ray makes an angle $$\theta_i$$ with the vertical in water, and then refracts at $$C$$ as it transitions into the air, and continues at angle $$\theta_r$$ with the vertical. From the Snell's law, $$\sin{\theta_r}=n\sin{\theta_i}$$ (1) (refractive index of an air is approximately unity). Let us extend the airborne segment of the ray back into the water, where it intersects line $$AB$$ at point $$D$$. As follows from the Snell's law, $$\theta_r > \theta_i$$, so point $$D$$ is positioned above point $$A$$. Above the water, the ray appears coming from point $$D$$ (please make the drawing, if you are actually working through this solution). Distance $$DB$$ is the perceived water depth. Distance $$AB$$ is the true water depth. From trigonometric relations for triangles $$ABC$$ and $$DBC$$, $$\tan {\theta_i}=BC/AB$$, (2) and $$\tan {\theta_r}=BC/DB$$. (3) Neglecting difference between $$\sin{\theta}$$ and $$\tan{\theta}$$ for small angles $$\theta$$, and substituting (2) and (3) into (1), we obtain from Snell's law: $$ AB=n \cdot DC$$. That is, the true depth of the pond is $$n$$ times its perceived depth, 1.33 m in our case.

### Subject: Physics (Newtonian Mechanics)

A bucket is suspended above a well from a cord winding around a drum. The drum weights $$M=15$$ kg. It is a uniform wooden cylinder of radius $$r=0.1$$ m which can rotate with negligible friction around the horizontal axis. The bucket weights $$m=5$$ kg. The bucket is released from the rest and falls into the well. Neglecting the weight of the cord, find the tension force in the cord while the bucket is falling.

Motion of three bodies has to be considered — that of the bucket, the drum, and the cord. (1) The bucket is in translational motion (when all its points move along parallel trajectories) along the vertical axis, which is fully determined by a trajectory of a single point — its centre of mass. The Newton’s 2-d law of motion for the bucket’s centre of mass states: $$ m \cdot \vec{a_b} = \vec{T_b} + m \vec{g} $$, where $$\vec{a_b}$$ is the bucket’s acceleration (we won’t differentiate anymore between the bucket and its centre of mass); $$\vec{T_b}$$ is the force exerted by cord directed up; $$\vec{g}$$ is the gravity acceleration directed down. (2) The drum rotates about a horizontal axis coming through its centre of mass. The only uncompensated force acting on the drum is that exerted by the cord $$\vec{T_c}$$, applied at a horizontal distance $$r$$ from the axis and directed down. This force creates a torque with a magnitude $$rT_c$$ (hereafter, $$a$$ stands for a vertical projection of the vector $$\vec{a}$$), which is equal to the rate of change of the drum's angular momentum $$I \gamma$$, where $$I$$ is the drum's moment of inertia about the rotation axis, and $$\gamma$$ is the angular acceleration: $$I \gamma = rT_c$$. (3) By the Newton’s 3-d law, the vertical segment of the cord is pulled on the ends by oppositely directed forces $$-\vec{T_b}$$ and $$-\vec{T_c}$$. It would experience an infinitely large acceleration, if these forces were not in balance. Hence, $$\vec{T_b}+\vec{T_c}=0$$. (4) And finally, the bucket's acceleration must be equal to the tangential acceleration of a point on the rim of the drum directly above the bucket (which follows from the assumption that the cord does not stretch, so all points in its vertical segment, from the bucket to the rim of the drum, have the same velocity and the same acceleration). Combing (1)-(4), expressing the vector equations in the vertical projection, positive down, and using $$T=T_c=-T_b$$ for the tension in the cord, yields: $$ m \cdot a_b = - T + m g $$ (2-nd law for the bucket) $$I \gamma = rT$$ (moment equation for the drum) $$ a_b =r \gamma $$ (kinematic condition (4) ) Solving for $$T$$ yields: $$T=mgI/(I+mr^2)$$. The moment of inertia of the drum (a uniform cylinder) is $$I=Mr^2/2$$, so $$T= g mM/(M+2m)$$. Substituting numbers in units of SI gives $$T=3 g$$ ($$kg \cdot m/s^2$$).

### Subject: Calculus

A shape in $$xy$$-plane is given by $$ x^2+y^2+9 \le 6x+2y $$. Find the volume of the solid obtained by revolving the shape about the $$y$$-axis.

Let us re-write the inequality as $$ (x^2-6x+9)+(y^2-2y) \le 0$$, or $$(x-3)^2+(y-1)^2 \le 1$$. We see, that the shape is a disk of a unit radius centered at $$(3,1)$$. The sought solid of revolution is, therefore, a torus. Its volume is composed of very thin rings, swept by the shape’s elementary sub-areas $$dA=dxdy$$ rotating about $$y$$-axis. The sub-areal $$x$$-coordinate is the radius of the ring which the sub-area generates. Each sub-area sweeps a volume $$dV=2 \pi x \cdot dA$$, which is the volume of a very thin rod of a length $$ 2 \pi x$$ and a base area $$dA$$, bent to form a circle of radius $$x$$. The total volume can be found by integration over the shape $$ V=\int \int {dV} = \int \int {2 \pi x dA} = 2 \pi x_c A$$, where $$ x_c=(1/A) \int \int {x dA} $$ is the x-coordinate of the shape’s centroid, as well as the centroid’s distance to the axis of rotation. (We just proved Pappus's theorem for the volume generated by a revolving shape: this volume is the product of the shape’s area and the circumference of the circle traveled by the shape’s centroid. The axis of rotation should be outside the shape, though.) For our disk, its centroid coincides with its center: $$x_c=3$$. Revolving about $$y$$-axis, the centroid makes a circle with a circumference $$6 \pi$$. The disk’s area $$A=\pi$$, since the radius is a unit. The sought volume is then: $$V=6 \pi^2$$.

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