Question

Imagine that we release a rock of mass m (which is initially at rest) at the surface of a lake and measure its position and velocity as functions of time while it sinks. The rock moves under the influence of three forces: gravity, buoyancy, and viscous drag. Let y represent the vertical position of the sinking rock, with the surface of the lake at y -0, and positive y upwards The net force on the rock is F =-[m-mdisplaced where g is the acceleration of gravity, mdisplaced is the mass of the water displaced by the sinking rock, y is a unit vector that points up, and β is the (positive) coefficient of viscosity. (Note that g is a positive number.) If the rock is initially at rest, we can write this as a first order differential equation involving constants and the y component of velocity: dt displaced 8-Bv (a) The rock's terminal velocity v, corresponds to the velocity at which the net force on the rock is zero. Determine v, (which will be negative) in terms of the various constants in the above equation, then-by defining u so that u- Vy - Vi-rewrite the above differential equation in terms of u and any necessary constants. Be careful with signs! (b) The rock is released from rest at the surface of the water at t = 0. Solve for u, and then Vy, the rock's velocity as a function of time. You should express your answer in terms of m, mdisplaced g, and t. (c) Using your answer to (b), calculate the rock's position as a function of time, y(t). Please confirm that your solutions to (b) and (c) correspond to a rock at rest at the surface of the water at t = 0.

Answer 1

I have written the detailed solution. In case if you have any problem with the steps comment down. All the best.

The differential equation describing the motion is m dv_y/dt + (m - m_displaced)g + beta b_y = 0 Let m - m_displaced = M Then we get m dv_y/dt + beta v_y + Mg = 0 dv_y/dt = -(beta/m v_y + M/m g) dv_y/beta/m v_y + M/m g integral dv_y/beta/m v_y + M/m g = -integral dt m/beta integral dv_y/v_y + M/beta g = -integral dt m/beta ln(v_y + M/beta g) = -t + c v_y + M/beta g = e^beta/m c e^beta/m t v_y = e^beta/m c e^beta/m t - M/beta g At time t = 0 the rock is at rest. So we get e^beta/m c = M/beta g c = m/beta ln(M/beta g) = m/beta ln(m - m_displaced/beta g) Now v_y = (m - m_displaced/beta g) e^- beta/m t - (m - m_displaced/beta g) v_y = (m - m_displaced/beta g)(e^- beta/m - 1) When the net force on the particle becomes zero, then the particle moves with a terminal velocity. So we get