A simple model for the gravity-driven motion of a glacier is to picture a large

Question

A simple model for the gravity-driven motion of a glacier is to picture a large block of ice, sliding down a rock incline of angle theta, where a thin layer of melt-water acts as a lubricant between the ice-block and rock face. See sketch on blackboard. Consider first the case where the glacier slides with a constant (terminal) velocity, VT, down the rock face. If the mass of the glacier is M, the viscosity of the water in the melt-water layer is mu, the water layer thickness is (constant and equal to) d, and the area of contact between the glacier and rock is A, what is VT? Express in terms of M, g (=gravitational acceleration), A, d, theta, and mu. In the case where the glacier is accelerating, write down the appropriate equation of motion for the glacier, i e., apply Newton's second law to the glacier. Let the instantaneous velocity of the glacier be V(t) and express your answer in terms V(t) and all of the parameters listed in part a). Using the equation of motion derived in part b) determine a detailed expression for V(t), assuming that the glacier starts moving from a state of rest.

Explanation / Answer

a) Using the equation:

F = A(v/d)

where F is force, is viscosity, A is area, v is velocity, and d is the distance of seperation. At terminal velocity, the force of the viscous fluid is equal to the force of gravity pulling the glacier down the hill, so:

sin()mg = A(vt/d)

Solving for the velocity:

vt = (d*sin()mg)/(A)

b) In the case that the glacier has not reached terminal velocity, this results in a differential equation in the form:

F = ma = dV/dt*m = sin()mg - A(V(t)/d)

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