A physics students is being chased downhill by a large cylindrical log of radius

Question

A physics students is being chased downhill by a large cylindrical log of radius 1.0-m that has an initial velocity of 25 m/s. If the hill has a height of 75-m, how fast will the log be moving when if reaches the bottom of the hill? If the student is initially 20-m from the top of the hill and is running with a speed of 5 m/s, can he/she make it to the bottom of the hill before being crushed by the log? If not how far down the hill will the student be able to run before being struck by the log? Assume the log is solid and of uniform density. Furthermore, assume that it rolls without slipping.

Explanation / Answer

radius of log, R = 1 m
initial velocity, u = 25 m/s
height of hill, h = 75 m

so moment of inertia of log = mR^2 [ where m is mass of log]
so at any point on the hill, if its velocity is v
rotational KE = 0.5*Iw^2 = 0.5*mR^2*(v^2/R^2) = 0.5mv^2
KE = 0.5mv^2
Net KE = mv^2

so, initial Total energy = mu^2 + mgh
final total energy = mv^2
from conservation of energy
u^2 + gh = v^2
v = sqroot(25^2 + 8.81*75) = 36.88 m/s
so the log is moving at 36.88 m/s at the bottom of the hill

initial distance from the top of the hill, di = 20 m
INITIAL SPEED OF STUDENT, U = 5 m/s
so, at 5 m/s, assuming angle of hill with the horizontal to be theta
u = (75/sin(theta) - 20)/t
t = [75/sin(theta) - 20]/5 = 25/sin(theta) - 4

but for the log
u = 25 m/s
v = 36.88 m/s
acceleration = a
distnace = 75/sin(theta)
so, 2*a*75/sin(theta) = v^2 - u^2 = 735.1344
a/sin(theta) = 4.900896

so, using v = u + at
t = (v - u)/a = (36.88 - 25)/4.900896sin(theta) = 2.424/sin(theta)

so, 25/sin(theta) - 4 = 2.424/sin(theta)
sin(theta) = 5.64
this is not possible hence the man cannot outrun the log of wood

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