The Inclined Plane motion problem is a classic problem in demonstrating the laws of motion. We've already provided the animation of a Block Sliding on Inclined Plane and here we will have a roller instead of the block. This necessitates the use of torque vectors as well as force vectors. It requires integration of the angular acceleration induced by torque, to obtain the current angle, `phi`, and current roller center displacement, `s`. Here we will shorten the phrase "inclined plane" to the term "ramp". The gravity force component that is perpendicular to the ramp surface results in a backward friction force that is large enough to avoid any slippage at the roller contact point. On the other hand, a block must slip in order to move but its friction is slider variable.
This page shows how the parallel and perpendicular force vectors sum together to be equivalent to the downward gravity force vector, `mg`. To sum two vectors graphically, we place the tail of an arrow representing the second vector at the tip of the arrow representing the first vector. Here the parallel and perpendicular force vectors are blue and they add up to the red downward gravity force vector as they must. The green arrow that represents the roller friction force vector at the contact point is the same length as the parallel force vector but points in the opposite direction.
The forces on the roller are:
`F_("parallel")=Mg sin theta \ \ \ \("gravity force along ramp")`
`F_("perpendicular")=Mg cos theta \ \ \ \("gravity force perpendicular to ramp")`
`F_("friction")= -Mgsin theta\ \ \ \ \("roller friction force")`
`vec tau=vec r times vec F \ \ \ \ \("torque that causes angular acceleration")`
`vec L=m vec r times vec v`
where `m` is the particle mass, `vec r` is its distance from its rotation axis, and `vec v` is its linear speed. The `times` symbol here denotes the cross product of two vectors. If the vectors are at angle `psi` with respect top each other, then the magnitude of the cross product is `|vec r| |vec v| sin psi`, where `psi` is the angle between the vectors and a vector expressed as `|vec a|` represents the magnitude of `vec a`. The direction of `vec L` is mutually perpendicular to each vector. Similarly to linear momentum, angular momentum is constant unless an outside force is applied. Now if we have a rigid body like our roller which is composed of many mass annuli at various radii `r`, then we can describe the angular momentum as:`vec L= I vec ((dphi)/dt)`
where `I` is called the moment of inertia and vec ((dphi)/dt)` is the rotation rate of the body and it is a vector pointing along the axis of rotation.The moment of inertia requires explanation: It is the inertia of a body to acceleration of its rotation angle. To accelerate the masses of a cylinder about its axis, requires that the masses in annuli at various radii from the axis accelerate. The mass of the shell between `r -(delta r)/2` and `r+(delta r)/2` is
`delta m=2pi r delta r l rho_("roller")`
where `l` is the length of the cylinder and `rho_("roller")` is the density of the cylinder. The inertia to angular acceleration of the annular mass, `m`, at radius r is`delta m r^2 vec (d^2phi)/dt^2= vec tau =vec R times vec F_("parallel") \ \ \ \ \(1)`
To obtain the moment of inertia we must integrate `delta m r^2` with respect to `delta r`. The result is`I=2pirho_("roller") lint_0^Rr'^3dr'= pi l rho_("roller") R^4/2`
But the mass of our cylinder is`M=pi l R^2 rho_("roller")`
so we can write:`I=(MR^2)/2`
The equations of motion are tricky because the same force, `Mg sin theta`, must accelerate both the angular rotation and the linear center of mass (CM) motion. So we have to back up a bit from our expression equation 1 for angular acceleration. By dividing out the radius from the torque expression in equation 1 the magnitude becomes force. And, of course, the expression for linear acceleration of the `CM` is `MR(d^2phi)/(dt^2)=F` where `F` is a fraction of `Mg sin theta. Then, using the sum of the two inertia terms, we can write the following expressions
`I/R (d^2phi)/dt^2+MR(d^2phi)(dt^2)= Mg sin theta\ \ \ \ (a)`
`MR/2 (d^2phi)/dt^2+MR(d^2phi)(dt^2)= Mg sin theta\ \ \ \ ("re-write of equation a") `
`(d^2phi)/dt^2= 2/(3MR) Mg sin theta\ \ \ \ ("equation for acceleration")`
`(ds)/dt=R (dphi)/dt=2/(3M) Mg sin thetat\ \ \ \ \("speed of the CM on ramp")`
`phi=1/(3MR) Mg sin thetat^2\ \ \ \ \ \("solution for phi")`
`s=Rphi=1/(3M) Mg sin thetat^2=g(t^2/3) sin theta\ \ \ \ \("displacement of the CM on ramp")`
`phi=s/R=g(t^2/(3R)) sin theta
We know that kinetic energy of a mass moving at speed v is `KE=1/2mv^2`. In our case the speed is `v_t=r(dphi)/dt` where `v_t` is the tangential speed of a mass at radius `r` from the axis. As discussed before, the incremental mass in a shell at radius r and of thickness `deltar` is
`delta m=2pi r delta r l rho_("roller")`
So our expression for the incremental kinetic energy at radius `r` is:`delta KE=1/2 2pi r delta r l rho_("roller")r^2((dphi)/dt)^2\ \ \ \ (KE increment) `
To get the rotating kinetic energy we integrate this expression with respect to r:`KE=int_0^R pi r l rho_("roller")r^2((dphi)/dt)^2dr=pi/4R^4lrho_("roller")((dphi)/dt)^2`
But the mass of the cylinder is `m=piR^2lrho_("roller") so our expression for the rotational kinetic energy simplifies to:`KE_("rotational")=1/4MR^2((dphi)/dt)^2=1/2 I ((dphi)/dt)^2`
The cylinder rolls without slipping down the ramp at rotation rate `(dphi)/dt` so its center of mass (axis) speed is `R(dphi)/dt` Then the kinetic energy associated with its center of mass (CM) motion is
`KE_("CM")=1/2MR^2((dphi)/dt)^2`
which is twice as large as `KE_("rotational")`. The roller's total kinetic energy is then:`KE_("total")=KE_("CM")+KE_("rotational")=3/4MR^2((dphi)/dt)^2`
`delta U`=-Mgs sintheta`
If we set `KE_("total")` equal to the negative of the change in potential energy, we can obtain `(dphi)/dt` directly from the kinetic energy:
`3/4MR^2((dphi)/dt)^2=-deltaU=Mgs sintheta`
`(ds')/dt=R(dphi)/dt=-sqrt(4/3sgsin theta)`
`(ds')/sqrt(s')=sqrt(4/3(gsin theta)dt`
integrating over s' and t we obtain:
`int_0^s (ds')/sqrt(s')=-2sqrt(s)=-sqrt(4/3gsin theta)int_0^t dt'=-tsqrt(4/3gsin(theta))`
solving for s we have:
`s=g(t^2/3) sin theta `
which agrees with the result from the equations of motion above.
The roller introduces a new concept to the equations of motion: Angular momentum of a rigid body. Angular momentum is defined for point particles but is usually applied to rigid extended bodies like our cylindrical roller. The roller has both rotational inertia about its axis and linear inertia of its mass. Linear inertia is usually defined for its center of mass which for a roller is its axis. The combination of these inertias results in acceleration of the roller at 2/3 the rate of a frictionless sliding block. Since it was convenient, I decided to include a block that slides down the ramp. Differing from a roller, the block usually experiences friction and therefore its total energy is not conserved. On the other hand, since the contact point of the roller is perfectly constrained (doesn't slip), it does conserve total energy so its motion can be computed by setting its kinetic energy gain equal to the loss of its potential energy. Note the lower speed of the roller because of its combined rotational and linear inertia.