Rigid body rolling along an inclined plane

[housemartin]

Homework Statement

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. y₀ - 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.

Homework Equations

conservation of mechanical energy:
1/2mv²+1/2Iw²+mgy = mgy₀
I = mK²
v = wR -> w = v/R
T = I*alfa

The Attempt at a Solution

using conservation of mechanical energy i get:
1/2mv²+1/2mK²v²/R²+mgy = mgy₀
from this velocity is:
v² = 2g(y₀-y)/(1+(K²/R²))
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)
K²alfa = R*g*sin(theta)
and if i use relation a = alfa*R, and multiply both sides of last equation by R, I get:
a = R²/K²*g*sin(theta)
to get velocity as a function of height, i rewrite a = dv/dt = dx/dt*(dv/dx) = v(dv/dx):
vdv = R²/K²*g*sin(theta)dx
if initial speed is zero and initial coordinate along the plane also zero, i get:
v²/2 = R²/K²*g*sin(theta)*x
and sin(theta)*x is just y₀ - y
so my final solution is bit different from true one:
v² = 2g(y₀-y)*R²/K²

So, where did I mistake?

 

[Doc Al]

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.

If the force of friction equaled the component of gravity parallel to the incline, then the net force would be zero and it wouldn't move down the incline.

You're right that friction is what makes it roll, but you'd need to figure out the friction force. Don't just assume it equals m*g*sin(theta), which it doesn't.

 

[housemartin]

so i should have write equation of motion like this:
ma = mg*sin(theta) - F_fric
now from here its really clear that if F_fric = mg*sin(theta) then a = 0.
Then if friction makes body roll, then i can write
torque = R*F_fric
I*alfa = R*F_fric
F_fric = I*alfa/R.
yes, now i get the correct answer. Thank you, I was struggling with this all morning

 

[Doc Al]

You got it now.

 

[rdl]

Your mistake is in assuming that the acceleration of the rigid body is solely due to the force of friction. In reality, there are two forces acting on the body as it rolls down the inclined plane: the force of gravity and the normal force from the plane. The normal force also contributes to the acceleration of the body, as it is responsible for keeping the body in contact with the plane and preventing it from slipping. Therefore, your equation for the torque should also include the normal force, and your final solution will be different.