Question

As a logging truck rounds a bend in the road, some logs come loose and begin to roll without slipping down the mountainside. The mountain slopes upward at 48.0 degrees above the horizontal, and we can model the logs as solid uniform cylinders of mass 575 kg and diameter 1.20m Find the acceleration of the logs as they roll down the mountain. What is the friction force on the rolling logs? Is it kinetic or static?

Answer 1

The logs are rolling means there is friction. Objects cannot roll down without friction, if there is no friction they slide down therefore the posted answer is is correct for logs sliding down without friction. So how do we solve the problem if there is friction and the logs are rolling down without slipping. Here we go: The torque of force of gravity with respect to the point where the log touches the ground is T = m g R sin theta. The moment of inertia of the log with respect to the same point is I = (3/2)(m)(R^2) NOTE: I = (1/2) (m)(R^2) is the moment of inertia of a cylinder with respect to its center. Now (angular acceleration) = (torque) / (moment of inertia). Finally: (linear acceleration ) = (angular acceleration)*(radius) ANSWER: a = 4.86 m/s^2
Good luck

Answer 2

Given that,
Angle between the horizontal and road θ=

Mass of the cylinder M=575 kg

Diameter of the cylinder d =1.20 m

then radius of cylinder 

a) Using parallel axis theorem,

the moment of inertia about the point of contact of the log with slope is 

the torque about the point of contact is τ=MgR sinθ

But we know that torque is τ=Iα

∴Iα=MgR sinθ

Angular acceleration α=

therefore acceleration a=αR

                                   a= (2gsin θ)/3

                                      =(2*9.8*sin 48)/3

                                      =4.8552

b) Since friction force provides the torque  to produce the angular acceleration about the center of the  mass

here f= frictional force

Moment of inertia about the center of the mass 

f=[( 575)*(9.8)*sin 48]/3=1395.8 N

c) it is a static friction

Answer 3

a) acceleration = - g sinθ / ( 1 +I_com/MR2 )
for uniform cylinder Icom =0.5MR2
∴ acceleration = - 9.8*sin48 / ( 1 + 0.5) = - 4.855 m/s2 ( in the negativedirection of the x - axis )

b) frictional froce is static.
∴ fs = - I_com a/ R2

=- 0.5Ma

= 1395N