Maximum angle of rolling without slipping.

[synth]

New here, first post.

Ok, a block of a certain material begins to slide on an inclined plane when the plane is inclined to a certain angle, theta, giving us tan(theta) = the coefficient of static friction. Now, if a solid cylinder is made out of the material, how would one go about finding the maximum angle that the cylinder would roll without slipping?

Thanks for help in advance.

 

[airbuzz]

you start to slip when the force, that is $$m\,g\,cos\vartheta$$ beats the static friction force that is $$m\,g\,sin\vartheta\mu$$, where $$\mu$$ is the static friction coefficient
If theta is little you roll without slipping, when theta becomes bigger than this value you begin to roll and slip together.
Bye

 

[ehild]

New here, first post.

Ok, a block of a certain material begins to slide on an inclined plane when the plane is inclined to a certain angle, theta, giving us tan(theta) = the coefficient of static friction. Now, if a solid cylinder is made out of the material, how would one go about finding the maximum angle that the cylinder would roll without slipping?

Thanks for help in advance.

If the cylinder rolls without slipping the velocity of its centre of mass is the same v as that of a point of its perimeter, that is

$$V_{CM}=v=r\omega$$.

The motion of the cylinder is composed of a translation of its CM under th influence of all forces, and a rotation around CM, under the influence of the sum of all torques. You have two equations, one for the translation and one for the rotation. Three forces act on the cylinder: gravity, (mg) normal force, (N) static friction, (Ffr). The equation of motion for the translation along a slope of inclination alpha is

$$m*a = mg\sin{\alpha} - F_{fr}$$.

The friction decelerates the translation, but it accelerates rolling. The torque of the static friction with respect to the axis through the centre of mass is

$$M=rF_{fr} = I\beta$$,

beta is the angular acceleration and I is the moment of inertia. For a solid cylinder,

$$I=\frac{1}{2}mr^2$$.

When it is pure rolling

$$a=r\beta =r^2F_{fr}/I$$.


Plugging this into the eq. for translation you get:

$$mr^2F_{fr}/I=mg\sin{\alpha}-F_{fr}$$

$$F_{fr}(1+\frac{mr^2}{I})= mg\sin{\alpha}$$

mr^2/I = 2 for the solid cylinder, so

$$F_{fr} =\frac{ mg\sin{\alpha}}{3}$$

$$F_{fr} \leq \mu N =\tan{\theta}*mg\cos{\alpha}$$

$$F_{fr} =\frac{ mg\sin{\alpha}}{3} \leq \tan{\theta}*mg\cos{\alpha} \rightarrow \tan{\alpha}\leq 3\tan(\theta)$$

In order to pure rolling, the tangent of the angle must not exceed 3 times of the tangent of the angle at which the block just starts to move.

ehild

 

[nanyzz]

but i got the answer is arctan 3*miu...
how come?