From Icm

Posted in Aircraft Model by admin on November 16, 2008

A cylinder (r = 0.139 m, Icm = 2.596E-2 kg ·m2, M = 1.69 kg) starts from rest and rolls without slipping down

...a plane with an angle of inclination of q = 21.9°. Find the time it takes it to travel 1.61 m along the incline.

Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline

SFx = m*a = m*g*sin(q) - Fs

where m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque

ST = I*a/r = r*Fs

so that we have for the statice friction force

Fs = I*a/r^2.

Plugging this into the first equation

m*a = m*g*sin(q) - I*a/r^2

and solving for the acceleration

a = m*g*sin(q)/(m + I/r^2)

Now, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is

L = 0.5*a*t^2

so that

t = sqrt(2*L/a)

Also, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by

2*pi*L/(2*pi*r) = 0.5*alpha*t^2

where alpha is the angular acceleration and is equal to a/r.

Conservation of energy can be employed as well

m*g*h = 0.5*m*v^2 + I*w^2

where h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields

v = sqrt(2*m*g*L*sin(q)/(m + I/r^2))

which is equal to the acceleration times the time traveled. From this

t = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a

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