(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

So rigid body (ring, cylinder, sphere), who's radius of gyration is known, rolls down an inclined plane without slipping. Find velocity as a function of height.

m - mass of the body. y_{0}- initial height, g - gravitational acceleration, v - speed. I - moment of inertia, K - radius of gyration. T - torque, a - linear acceleration, alfa - angular acceleration, w - angular speed. theta - inclined plane angle with horizontal.

2. Relevant equations

conservation of mechanical energy:

1/2mv^{2}+1/2Iw^{2}+mgy = mgy_{0}

I = mK^{2}

v = wR -> w = v/R

T = I*alfa

3. The attempt at a solution

using conservation of mechanical energy i get:

1/2mv^{2}+1/2mK^{2}v^{2}/R^{2}+mgy = mgy_{0}

from this velocity is:

v^{2}= 2g(y_{0}-y)/(1+(K^{2}/R^{2}))

which is correct answer as my textbook says.

However I get in trouble trying do this problem in different way - using forces. As i think, if body rolls without slipping that means that force of friction is equal to the m*g*sin(theta), only friction makes this body roll. All torques about center of mass except that of friction equal to zero, and I get:

T = R*m*g*sin(theta)

I*alfa = R*m*g*sin(theta)

K^{2}alfa = R*g*sin(theta)

and if i use relation a = alfa*R, and multiply both sides of last equation by R, I get:

a = R^{2}/K^{2}*g*sin(theta)

to get velocity as a function of height, i rewrite a = dv/dt = dx/dt*(dv/dx) = v(dv/dx):

vdv = R^{2}/K^{2}*g*sin(theta)dx

if initial speed is zero and initial coordinate along the plane also zero, i get:

v^{2}/2 = R^{2}/K^{2}*g*sin(theta)*x

and sin(theta)*x is just y_{0}- y

so my final solution is bit different from true one:

v^{2}= 2g(y_{0}-y)*R^{2}/K^{2}

So, where did I mistake?

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# Homework Help: Rigid body rolling along an inclined plane

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