Physics · Ch 1 — Rotational Dynamics

Linear Acceleration and Speed While Pure Rolling Down an Inclined Plane

1.11.1

Linear Acceleration and Speed While Pure Rolling Down an Inclined Plane

Figure 1.18Fig. 1.18: Rolling along an incline — a body rolling without slipping down an incline of angle θ, travelling distance s along the plane while falling through vertical height h = s sin θ

Fig. 1.18 — Fig. 1.18: Rolling along an incline — a body rolling without slipping down an incline of angle θ, travelling distance s along the plane while falling through vertical height h = s sin θ

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A rigid circular object (cylinder, disc or sphere) of mass M and radius R shown at the top of an inclined plane that makes angle $\theta$ with the horizontal, about to roll down without slipping. The object's centre is marked, with the incline's surface drawn as a straight ramp rising at angle $\theta$ from the horizontal ground; the object's radius R is marked from its centre to the point of contact with the incline surface. The linear distance to be travelled ALONG the incline is marked as s, related to the vertical height fallen h by $s=h/\sin\theta$ (i.e. $h=s\sin\theta$ ), setting up the energy-conservation calculation of this section that relates the vertical drop h …

Consider a rigid object of mass M and radius R, with radius of gyration K, released from rest at the top of an inclined plane of angle $\theta$ , and allowed to roll DOWN without slipping. As it descends, its gravitational potential energy converts entirely into the kinetic energy of rolling (Eq. 1.18), since static friction (doing no net work, as there is no relative sliding at the contact point) does not dissipate any energy here. If the object falls through a vertical height h while starting from rest, energy conservation gives $Mgh=\frac{1}{2}Mv^2\left(1+\frac{K^2}{R^2}\right) \quad\Rightarrow\quad v^2=\frac{2gh}{1+K^2/R^2} \quad\Rightarrow\quad v=\sqrt{\dfrac{2gh}{1+K^2/R^2}} \qquad \text{--- (1.19)}$

The linear distance actually travelled ALONG the incline while falling through height h is $s=\dfrac{h}{\sin\theta}$ (equivalently $h=s\sin\theta$ ). Since the object starts from rest ( $u=0$ ) and reaches speed v (from Eq. 1.19) after travelling this distance s with some constant linear acceleration a along the incline, the ordinary kinematic relation $v^2=u^2+2as$ gives $a=\frac{v^2}{2s}=\frac{1}{2s}\cdot\frac{2gh}{1+K^2/R^2}=\frac{gh}{s(1+K^2/R^2)}=\frac{g\sin\theta}{1+K^2/R^2} \qquad \text{--- (1.20)}$ (using $h/s=\sin\theta$ in the last step).

For comparison, an object sliding down the SAME frictionless incline (no rotation at all) would have acceleration exactly $g\sin\theta$ and would reach speed exactly $\sqrt{2gh}$ -- so BOTH the rolling speed and the rolling acceleration are reduced from their pure-sliding values by the identical factor $\left(1+\dfrac{K^2}{R^2}\right)^{-1}$ . This factor depends ONLY on the ratio $K^2/R^2$ -- i.e. purely on the object's SHAPE -- and not at all on its mass or its actual size, which is exactly why a solid sphere, a hollow sphere, a solid cylinder and a ring, released together from the same height on the same incline, reach the bottom in a definite, mass-and-size-independent order (fastest to slowest) determined purely by their shape. …

Table Table 1.1Table 1.1: Analogous kinematical equations (ω₀ is the initial angular velocity)

Equation for translational motion	Analogous equation for rotational motion
$v_{av} = \dfrac{u+v}{2}$	$\omega_{av} = \dfrac{\omega_0+\omega}{2}$
$a = \dfrac{dv}{dt} = \dfrac{v-u}{t}$ $\therefore v = u+at$	$\alpha = \dfrac{d\omega}{dt} = \dfrac{\omega-\omega_0}{t}$ $\therefore \omega = \omega_0+\alpha t$

Table Table 2Table 2: Analogous quantities between translational motion and rotational motion

Translational motion: Quantity	Symbol/expression	Rotational motion: Quantity	Symbol/expression	Inter-relation, if possible
Linear displacement	$\vec{s}$	Angular displacement	$\vec{\theta}$	$\vec{s} = \vec{\theta} \times \vec{r}$
Linear velocity	$\vec{v} = \dfrac{d\vec{s}}{dt}$	Angular velocity	$\vec{\omega} = \dfrac{d\vec{\theta}}{dt}$	$\vec{v} = \vec{\omega} \times \vec{r}$
Linear acceleration	$\vec{a} = \dfrac{d\vec{v}}{dt}$	Angular acceleration	$\vec{\alpha} = \dfrac{d\vec{\omega}}{dt}$	$\vec{\alpha} = \vec{a} \times \vec{r}$
Inertia or mass	$m$	Rotational inertia or moment of inertia	$I$	$I = \int r^2 dm = \sum m_i r_i^2$
Linear momentum	$\vec{p} = m\vec{v}$	Angular momentum	$\vec{L} = I\vec{\omega}$	$\vec{L} = \vec{r} \times \vec{p}$
Force	$\vec{f} = \dfrac{d\vec{p}}{dt}$	Torque	$\vec{\tau} = \dfrac{d\vec{L}}{dt}$	$\vec{\tau} = \vec{r} \times \vec{f}$
Work	$W = \vec{f} \cdot \vec{s}$	Work	$W = \vec{\tau} \cdot \vec{\theta}$	——
Power	$P = \dfrac{dW}{dt} = \vec{f} \cdot \vec{v}$	Power	$P = \dfrac{dW}{dt} = \vec{\tau} \cdot \vec{\omega}$	——

Table Table 3Table 3: Expressions for moment of inertias for some symmetric objects

Object	Axis	Expression of moment of inertia	Figure
Thin ring or hollow cylinder	Central	$I = MR^2$	Thin ring or hollow cylinder — central axis
Thin ring	Diameter	$I = \dfrac{1}{2} MR^2$	Thin ring — diameter axis
Annular ring or thick walled hollow cylinder	Central	$I = \dfrac{1}{2} M\left(r_2^2+r_1^2\right)$	Annular ring or thick walled hollow cylinder — central axis
Uniform disc or solid cylinder	Central	$I = \dfrac{1}{2} MR^2$	Uniform disc or solid cylinder — central axis
Uniform disc	Diameter	$I = \dfrac{1}{4} MR^2$	Uniform disc — diameter axis
Thin walled hollow sphere	Central	$I = \dfrac{2}{3} MR^2$	Thin walled hollow sphere — central axis
Solid sphere	Central	$I = \dfrac{2}{5} MR^2$	Solid sphere — central axis
Uniform symmetric spherical shell	Central	$I = \dfrac{2}{5} M \dfrac{\left(r_2^5-r_1^5\right)}{\left(r_2^3-r_1^3\right)}$	Uniform symmetric spherical shell — central axis
Thin uniform rod or rectangular plate	Perpendicular to length and passing through centre	$I = \dfrac{1}{12} ML^2$	Thin uniform rod or rectangular plate — perpendicular to length and passing through centre axis
Thin uniform rod or rectangular plate	Perpendicular to length and about one end	$I = \dfrac{1}{3} ML^2$	Thin uniform rod or rectangular plate — perpendicular to length and about one end axis
Uniform plate or rectangular parallelepiped	Central	$I = \dfrac{1}{12} M(L^2+b^2)$	Uniform plate or rectangular parallelepiped — central axis
Uniform solid right circular cone	Central	$I = \dfrac{3}{10} MR^2$	Uniform solid right circular cone — central axis

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