Question

The box has a mass m and slides down the smooth chute having the shape of a parabola. If it has an initial velocity of v0 at the origin determine its velocity as a function of x. Also, what is the normal force on the box, and the tangential acceleration as a function of x?

Answer 1

`v = \sqrt{v_(o)^(2)+2\cdot g\cdot a\cdot (x_(o)^(2)-x^(2))}`,
`N = m\cdot \left[\frac{v_(o)^(2)+2\cdot g \cdot a \cdot (x_(o)^(2)-x^(2))}{\frac{\left[1 + 4\cdot a^(2)\cdot x^(2) \right]^{(3)/(2) }}a } + \frac{g}{\sqrt{1+4\cdot a^(2)\cdot x^(2)}} \right]`,
`a = g \cdot \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^(2)\cdot x^(2)}}`

Step-by-step explanation:

An expression for the velocity can be derived by applying the Principle of Energy Conservation:

`(1)/(2)\cdot m \cdot v_(o)^(2) + m\cdot g \cdot y_(o) = (1)/(2)\cdot m \cdot v^(2) + m\cdot g \cdot y`

`(1)/(2) \cdot v_(o)^(2) + g \cdot y_(o) = (1)/(2) \cdot v^(2) + g \cdot y`

`v = \sqrt{v_(o)^(2)+2\cdot g\cdot (y_(o)-y)}`

Let assume that chute is frictionless, so that box is moved because of gravity. The equations of equilibrium for the box are:

`\Sigma F_(x') = m\cdot g \cdot \sin \theta = m \cdot a\\\Sigma F_(y') = N - m\cdot g \cdot \cos \theta = m\cdot (v^(2))/(\rho)`

The expression for the radius of curvature and trigonometric functions are, respectively:

`\sin \theta = (dy)/(ds) = \frac{1}{\sqrt{1+((dy)/(dx) )^(2)} }\cdot (dy)/(dx)`

`\cos \theta = (dx)/(ds) = \frac{1}{\sqrt{1+((dy)/(dx) )^(2)}}`

`\rho = \frac{\left[1 + ((dy)/(dx) )^(2) \right]^{(3)/(2)} }`

Let assume that chute has the following form:

`y = a\cdot x^(2)`

First and second derivatives of the function are, respectively:

`(dy)/(dx)= 2\cdot a \cdot x`

`(d^(2)y)/(dx^(2)) = 2\cdot a`

Trigonometric functions and radius of curvature are:

`\sin \theta = \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^(2)\cdot x^(2)} }`

`\cos \theta = \frac{1}{\sqrt{1+4\cdot a^(2)\cdot x^(2)} }`

`\rho = \frac{\left[1+4\cdot a^(2)\cdot x^(2)\right]^{(3)/(2) }}2\cdot`

The velocity as a function of x is:

`v = \sqrt{v_(o)^(2)+2\cdot g\cdot a\cdot (x_(o)^(2)-x^(2))}`

The normal force on the box as a function of x is:

`N = m \cdot \left[ (v^(2))/(\rho) + g\cdot \cos \theta \right]`

`N = m\cdot \left[\frac{v_(o)^(2)+2\cdot g \cdot a \cdot (x_(o)^(2)-x^(2))}{\frac{\left[1 + 4\cdot a^(2)\cdot x^(2) \right]^{(3)/(2) }}a } + \frac{g}{\sqrt{1+4\cdot a^(2)\cdot x^(2)}} \right]`

The tangential acceleration on the box as a function of x is:

`a = g\cdot \sin \theta`

`a = g \cdot \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^(2)\cdot x^(2)}}`