Question

Worked out solution

Problem 2/31 2/45 The cone falling with a speed vo strikes and pene- trates the block of packing material. The accelera- tion of the cone after impact is a g c2, where c is a positive constant and y is the penetration dis- tance. If the maximum penetration depth is ob- served to be ym, determine the constant c. Uo Problem 2/45

Answer 1

we know that acceleration is given by:

a = V*(dV/dy)

V*dV = a*dy

Now given that

a = g - cy^2

V_initial = initial velocity = v0

V_final = final velocity = 0 m/sec

max penetration depth = y_max = ym

And we need to find c = ?

So Using above equation:

V*dV = a*dy

Integrating both sides

(limit for velocity is v = v0 to v = 0 & limit for depth is y = 0 to y = ym)

$\int_{v_{0}}^{0}v*dv = \int_{0}^{y_{m}}a*dy$

$\int_{v_{0}}^{0}v*dv = \int_{0}^{y_{m}}(g - cy^2)*dy$

$[\frac{v^2}{2}]_{v_{0}}^{0} = [gy - \frac{cy^3}{3}]_{0}^{y_{m}}$

$[\frac{0^2}{2} - (\frac{v_{0}^2}{2})]= [(gy_{m} - \frac{c(y_{m})^3}{3}) - (g*0 - \frac{c(0)^3}{3})]$

$\frac{-v_{0}^2}{2} = gy_{m} - \frac{cy_{m}^3}{3}$

$\frac{cy_{m}^3}{3} = gy_{m} + \frac{v_{0}^2}{2}$

$cy_{m}^3 = 3gy_{m} + \frac{3v_{0}^2}{2}$

$c = \frac{3gy_{m}}{y_{m}^3} + \frac{3v_{0}^2}{2y_{m}^3}$

$c = \frac{6gy_{m}}{2y_{m}^3} + \frac{3v_{0}^2}{2y_{m}^3}$

$c = \frac{3}{2y_{m}^3}[2gy_{m} + v_{0}^2]$

Please Upvote.

Comment below if you have any query.