A 5000 kg boulder with radius 1 m is rolling down a 10 m tall hill with a 15° gr
ID: 1613911 • Letter: A
Question
A 5000 kg boulder with radius 1 m is rolling down a 10 m tall hill with a 15° grade (that’s the angle the hill makes with horizontal), chasing Indiana Jones’ jeep. The body of the jeep has a mass of 1800 kg, and its four wheels are 10 kg each and have radii of 0.5 m. If Indy’s jeep has a head start of 4 seconds, will he outrun the boulder to the bottom of the hill? Treat the boulder as a solid sphere and the tires as solid disks. Assume that everything starts from rest, and also assume that no slipping occurs.
Explanation / Answer
for the boulder :
m = mass = 5000 kg
r = radius = 1 m
I = moment of inertia = (0.4) m r2
force equation parallel to incline is given as
mgSin15 - f = ma
f = mgSin15 - ma
torque equation is given as
f r = I a/r
f = (0.4) m r2 a/r2
f = (0.4) ma
(0.4) ma = mgSin15 - ma
a = g Sin15/1.4
a = 1.81 m/s2
L = distance travelled = h /Sin15 = 10/Sin15 = 38.5 m
t = time taken
using the equation
L = (0.5) a t2
38.5 = (0.5) (1.81) t2
t = 6.52 sec
for the jeep :
consider the motion of one wheel
m = mass = 10 kg
r = radius = 0.5 m
I = moment of inertia = (0.5) m r2
force equation parallel to incline is given as
mgSin15 - f = ma
f = mgSin15 - ma
torque equation is given as
f r = I a/r
f = (0.5) m r2 (a/r2 )
f = (0.5) ma
(0.5) ma = mgSin15 - ma
a = g Sin15/1.5
a = 1.69 m/s2
L = distance travelled = h /Sin15 = 10/Sin15 = 38.5 m
t = time taken
using the equation
L = (0.5) a t2
38.5 = (0.5) (1.69) t2
t = 6.75 sec
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