Concept Card: Rolling Motion – Combined Rotation and Translation & Pure Rolling
1. Kinematic Decomposition of Rolling Motion:
The general plane motion of a symmetric rigid body (such as a cylinder, sphere, or wheel) rolling along a plane surface can be decomposed into two simultaneous, superposed motions:
- Pure Translation: The entire body translates with the instantaneous linear velocity of its centre of mass $\vec{v}_{cm}$ and linear acceleration $\vec{a}_{cm}$.
- Pure Rotation: The body rotates about a central axis passing through its centre of mass with instantaneous angular velocity $\vec{\omega}$ and angular acceleration $\vec{\alpha}$.
The instantaneous velocity $\vec{v}_P$ of any arbitrary particle $P$ at position $\vec{r}’$ relative to the $CM$ is given by the vector addition of these two components:
$$\mathbf{\vec{v}_P = \vec{v}_{cm} + \vec{v}_{P/cm} = \vec{v}_{cm} + (\vec{\omega} \times \vec{r}’)}$$
2. Condition for Pure Rolling (Rolling Without Slipping)
Pure rolling occurs when there is no relative slipping between the instantaneous contact point $C$ of the rolling body and the supporting surface at the contact interface:
Fundamental Condition for Rolling Without Slipping:
$$\vec{v}_{\text{contact point on body}} = \vec{v}_{\text{supporting surface}}$$
For a symmetric circular body of radius $R$ rolling along a stationary (fixed) horizontal surface:
$$v_C = v_{cm} – \omega R = 0 \implies \mathbf{v_{cm} = \omega R}$$
Differentiating with respect to time for pure rolling with acceleration:
$$\mathbf{a_{cm} = \alpha R}$$
Kinematic Slip Scenarios:
- Forward Slipping (Skidding / Over-speeding): $v_{cm} > \omega R$. The contact point slips forward ($v_C > 0$). Kinetic friction acts in the backward direction to reduce $v_{cm}$ and increase $\omega$ until pure rolling is restored.
- Backward Slipping (Spinning in Place): $v_{cm} < \omega R$. The contact point slips backward ($v_C < 0$). Kinetic friction acts in the forward direction to increase $v_{cm}$ and reduce $\omega$.
- Moving Support Surface: If the surface itself moves with velocity $\vec{v}_{\text{plank}}$ (e.g., rolling on a flatcar or plank), the condition becomes:
$$\vec{v}_{cm} + \vec{\omega} \times \vec{R} = \vec{v}_{\text{plank}}$$
3. Instantaneous Axis of Rotation (IAOR)
At any instant during pure rolling on a stationary surface, the contact point $C$ is momentarily at rest ($v_C = 0$). Consequently:
- The combined translation and rotation can be represented as pure rotation about an instantaneous axis passing through contact point $C$ with angular velocity $\omega$.
- The velocity of any point $P$ at distance $r_P$ from $C$ is simply directed perpendicular to the line segment joining $C$ to $P$, with magnitude:
$$v_P = \omega \cdot r_P$$ - Velocities of Key Geometric Points:
- Lowest point (Contact Point $C$): $r_C = 0 \implies \mathbf{v_C = 0}$.
- Centre of Mass ($CM$): $r_{cm} = R \implies \mathbf{v_{cm} = \omega R}$.
- Highest point (Top Rim $A$): $r_A = 2R \implies \mathbf{v_{\text{top}} = \omega (2R) = 2 v_{cm}}$ (directed horizontally forward).
- Equatorial rim points (Forward $B$ and Rear $D$): $r = \sqrt{R^2 + R^2} = R\sqrt{2} \implies \mathbf{v = \sqrt{2} v_{cm}}$ inclined at $45^\circ$ to the horizontal.
- Cycloidal Trajectory: Any point on the circumference traces out a cycloid in space. The instantaneous speed of a rim point as a function of the angle $\theta$ from the lowest point is:
$$v(\theta) = 2 v_{cm} \sin\left(\frac{\theta}{2}\right)$$
4. Total Kinetic Energy in Pure Rolling
By Koenig’s theorem for rigid body mechanics, the total kinetic energy of a rolling body is the sum of translational kinetic energy of the $CM$ and rotational kinetic energy about the $CM$:
$$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$$
In pure rolling, substitute $\omega = v_{cm}/R$ and write $I_{cm} = M k^2$ (where $k$ is the radius of gyration):
$$K = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} M k^2 \left(\frac{v_{cm}}{R}\right)^2 = \mathbf{\frac{1}{2} M v_{cm}^2 \left(1 + \frac{k^2}{R^2}\right)}$$
Equivalence via IAOR: Pure rotation about $C$ with $I_C = I_{cm} + M R^2 = M(k^2 + R^2)$ gives identically:
$$K = \frac{1}{2} I_C \omega^2 = \frac{1}{2} M R^2 \left(1 + \frac{k^2}{R^2}\right) \left(\frac{v_{cm}}{R}\right)^2 = \frac{1}{2} M v_{cm}^2 \left(1 + \frac{k^2}{R^2}\right)$$
| Body | $I_{cm}$ | $\frac{k^2}{R^2}$ | $K_{\text{trans}} / K_{\text{total}}$ | $K_{\text{rot}} / K_{\text{total}}$ |
|---|---|---|---|---|
| Thin Ring / Hollow Cylinder | $M R^2$ | $1$ | $\frac{1}{2} = 50\%$ | $\frac{1}{2} = 50\%$ |
| Uniform Disc / Solid Cylinder | $\frac{1}{2} M R^2$ | $\frac{1}{2} = 0.5$ | $\frac{2}{3} \approx 66.7\%$ | $\frac{1}{3} \approx 33.3\%$ |
| Hollow Sphere (Thin Shell) | $\frac{2}{3} M R^2$ | $\frac{2}{3} \approx 0.67$ | $\frac{3}{5} = 60\%$ | $\frac{2}{5} = 40\%$ |
| Uniform Solid Sphere | $\frac{2}{5} M R^2$ | $\frac{2}{5} = 0.4$ | $\frac{5}{7} \approx 71.4\%$ | $\frac{2}{7} \approx 28.6\%$ |
5. Dynamics of Pure Rolling Down an Inclined Plane
For a body of mass $M$, radius $R$, and radius of gyration $k$ released from rest on an incline of angle $\theta$:
- Linear Acceleration of Centre of Mass:
$$\mathbf{a_{cm} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}}$$ - Static Friction Force:
$$\mathbf{f_s = \frac{M g \sin\theta}{1 + \frac{R^2}{k^2}}}$$ - Minimum Friction Coefficient to Prevent Slipping:
$$\mathbf{\mu_{\min} = \frac{\tan\theta}{1 + \frac{R^2}{k^2}}}$$ - Speed at the Bottom (Vertical Drop $h$):
$$\mathbf{v = \sqrt{\frac{2 g h}{1 + \frac{k^2}{R^2}}}}$$ - The Incline Race: Smaller $\frac{k^2}{R^2}$ yields larger acceleration $a_{cm}$ and higher final speed $v$. Therefore:
$$\mathbf{\text{Solid Sphere } (0.40) > \text{Solid Disc } (0.50) > \text{Hollow Sphere } (0.67) > \text{Ring } (1.00)}$$
The solid sphere always wins the race down the incline!
6. Role of Friction in Pure Rolling
- Zero Work by Static Friction: Because the contact point $C$ has zero instantaneous velocity relative to the ground ($v_C = 0$), the power delivered by static friction is $P = \vec{f}_s \cdot \vec{v}_C = 0$. Hence:
$$W_{f_s} = \int \vec{f}_s \cdot d\vec{s}_C = \mathbf{0}$$
Mechanical energy is perfectly conserved during pure rolling on a stationary surface! - Direction of Static Friction: Static friction does NOT automatically oppose motion! Depending on where an external force is applied to a rolling body:
- If pulled at the top ($h = R$): $f_s$ acts forward (in the direction of applied force).
- If pulled at the critical height $h_0 = k^2/R$: $f_s = \mathbf{0}$ (no friction required!).
- If pulled through the center ($h = 0$): $f_s$ acts backward.
Solved Examples
Example 1 (Direct Conceptual Application – Velocity Field & IAOR in Pure Rolling):
A circular wheel of radius $R = 0.50\text{ m}$ rolls without slipping on a horizontal track with a uniform center-of-mass velocity $v_{cm} = 10.0\text{ m/s}$ directed to the right.
(a) Determine the angular velocity $\omega$ of the wheel.
(b) Calculate the magnitude and direction of the linear velocity of:
(i) Contact point $C$ at the bottom.
(ii) Topmost point $A$ on the rim.
(iii) Point $B$ on the leading horizontal edge of the rim.
(iv) Point $D$ located along the vertical radius midway between the center and the top rim ($0.25\text{ m}$ above the center).
(c) Demonstrate that the Instantaneous Axis of Rotation (IAOR) reproduces these exact velocities.
Solution:
(a) Angular Velocity:
For pure rolling on a stationary surface:
$$\omega = \frac{v_{cm}}{R} = \frac{10.0\text{ m/s}}{0.50\text{ m}} = \mathbf{20.0\text{ rad/s}} \quad (\text{clockwise})$$
(b) Velocity Analysis via Vector Superposition ($\vec{v} = \vec{v}_{cm} + \vec{\omega} \times \vec{r}’$):
Let $+x$ be to the right, $+y$ be vertically upward, and $\vec{\omega} = -20.0 \hat{k}\text{ rad/s}$.
$$\vec{v}_{cm} = 10.0 \hat{i}\text{ m/s}$$
– (i) Point $C$ (bottom): $\vec{r}’ = -R \hat{j} = -0.50 \hat{j}$.
$$\vec{v}_{C/cm} = (-20.0 \hat{k}) \times (-0.50 \hat{j}) = -10.0 \hat{i}$$
$$\vec{v}_C = 10.0 \hat{i} – 10.0 \hat{i} = \mathbf{\vec{0}} \quad (v_C = 0)$$
– (ii) Point $A$ (top): $\vec{r}’ = +0.50 \hat{j}$.
$$\vec{v}_{A/cm} = (-20.0 \hat{k}) \times (0.50 \hat{j}) = +10.0 \hat{i}$$
$$\vec{v}_A = 10.0 \hat{i} + 10.0 \hat{i} = \mathbf{20.0 \hat{i}\text{ m/s}} \quad (v_A = 2 v_{cm} = 20.0\text{ m/s})$$
– (iii) Point $B$ (front rim): $\vec{r}’ = +0.50 \hat{i}$.
$$\vec{v}_{B/cm} = (-20.0 \hat{k}) \times (0.50 \hat{i}) = -10.0 \hat{j}$$
$$\vec{v}_B = 10.0 \hat{i} – 10.0 \hat{j} \implies v_B = \sqrt{10.0^2 + (-10.0)^2} = \mathbf{10\sqrt{2}\text{ m/s}} \approx 14.14\text{ m/s}$$
Direction: $45^\circ$ below the horizontal.
– (iv) Point $D$ ($0.25\text{ m}$ above center): $\vec{r}’ = +0.25 \hat{j}$.
$$\vec{v}_{D/cm} = (-20.0 \hat{k}) \times (0.25 \hat{j}) = +5.0 \hat{i}$$
$$\vec{v}_D = 10.0 \hat{i} + 5.0 \hat{i} = \mathbf{15.0 \hat{i}\text{ m/s}} \quad (v_D = 15.0\text{ m/s})$$
(c) Verification via Instantaneous Axis of Rotation (IAOR at $C$):
The body rotates as a rigid unit about point $C$ with $\omega = 20.0\text{ rad/s}$:
– $v_C = \omega (0) = 0$.
– $v_A = \omega (r_{CA}) = (20.0)(2R) = (20.0)(1.0\text{ m}) = \mathbf{20.0\text{ m/s}}$.
– $v_B = \omega (r_{CB}) = \omega (R\sqrt{2}) = (20.0)(0.50\sqrt{2}) = \mathbf{10\sqrt{2}\text{ m/s}}$.
– $v_D = \omega (r_{CD}) = \omega (R + 0.25) = (20.0)(0.75\text{ m}) = \mathbf{15.0\text{ m/s}}$.
Takeaway: The IAOR method bypasses vector additions by translating the problem into pure rotation about the stationary contact point.
Example 2 (Mathematical Formulation – The Great Incline Race):
A uniform solid sphere, a solid cylinder (disc), and a thin spherical shell (hollow sphere), each having identical mass $M = 2.0\text{ kg}$ and outer radius $R = 0.20\text{ m}$, are released simultaneously from rest at the top of an inclined plane of angle $\theta = 30^\circ$ and length $L = 7.0\text{ m}$. All bodies roll without slipping. (Take $g = 9.8\text{ m/s}^2$).
(a) Derive the linear acceleration $a_{cm}$ and determine its numerical value for each body.
(b) Calculate the time taken by each body to reach the bottom and establish their arrival sequence.
(c) Calculate the final velocity $v$ of each body at the bottom of the incline.
(d) Calculate the static friction force $f_s$ acting on each body during descent.
Solution:
(a) Acceleration of Rolling Bodies:
Along the incline: $M g \sin\theta – f_s = M a_{cm}$.
Torque about $CM$: $f_s R = I_{cm} \alpha = (M k^2)(a_{cm}/R) \implies f_s = M a_{cm} (k^2/R^2)$.
$$M g \sin\theta – M a_{cm} \left(\frac{k^2}{R^2}\right) = M a_{cm} \implies \mathbf{a_{cm} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}}$$
Here $g \sin 30^\circ = (9.8)(0.50) = 4.90\text{ m/s}^2$.
– Solid Sphere: $\frac{k^2}{R^2} = \frac{2}{5} = 0.40 \implies a_{\text{sphere}} = \frac{4.90}{1 + 0.40} = \frac{4.90}{1.40} = \mathbf{3.50\text{ m/s}^2}$.
– Solid Cylinder: $\frac{k^2}{R^2} = \frac{1}{2} = 0.50 \implies a_{\text{cyl}} = \frac{4.90}{1 + 0.50} = \frac{4.90}{1.50} = \mathbf{3.267\text{ m/s}^2}$.
– Hollow Sphere: $\frac{k^2}{R^2} = \frac{2}{3} \approx 0.667 \implies a_{\text{hollow}} = \frac{4.90}{1 + \frac{2}{3}} = \frac{4.90 \times 3}{5} = \mathbf{2.94\text{ m/s}^2}$.
(b) Time of Descent ($L = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2L}{a}}$):
– Solid Sphere: $t_{\text{sphere}} = \sqrt{\frac{2(7.0)}{3.50}} = \sqrt{4.00} = \mathbf{2.00\text{ s}}$.
– Solid Cylinder: $t_{\text{cyl}} = \sqrt{\frac{2(7.0)}{3.267}} = \sqrt{4.286} = \mathbf{2.07\text{ s}}$.
– Hollow Sphere: $t_{\text{hollow}} = \sqrt{\frac{2(7.0)}{2.94}} = \sqrt{4.762} = \mathbf{2.18\text{ s}}$.
$$\mathbf{\text{Arrival Order: 1st Solid Sphere } (2.00\text{ s}) \rightarrow \text{2nd Solid Cylinder } (2.07\text{ s}) \rightarrow \text{3rd Hollow Sphere } (2.18\text{ s})}$$
(c) Final Speed at the Bottom ($v = \sqrt{2 a L}$):
– Solid Sphere: $v_{\text{sphere}} = \sqrt{2(3.50)(7.0)} = \sqrt{49.0} = \mathbf{7.00\text{ m/s}}$.
– Solid Cylinder: $v_{\text{cyl}} = \sqrt{2(3.267)(7.0)} = \sqrt{45.73} = \mathbf{6.76\text{ m/s}}$.
– Hollow Sphere: $v_{\text{hollow}} = \sqrt{2(2.94)(7.0)} = \sqrt{41.16} = \mathbf{6.42\text{ m/s}}$.
(d) Static Friction Force ($f_s = M a_{cm} \frac{k^2}{R^2}$):
– Solid Sphere: $f_s = (2.0)(3.50)(0.40) = \mathbf{2.80\text{ N}}$.
– Solid Cylinder: $f_s = (2.0)(3.267)(0.50) = \mathbf{3.27\text{ N}}$.
– Hollow Sphere: $f_s = (2.0)(2.94)\left(\frac{2}{3}\right) = \mathbf{3.92\text{ N}}$.
Takeaway: Mass and radius completely cancel out! The race depends strictly on the dimensionless shape factor $k^2/R^2$. Bodies with mass concentrated closer to the center possess smaller rotational inertia, leaving more gravitational energy for translational speed.
Example 3 (Multi-Concept Linkage – Horizontal Force on a Cylinder at Varying Heights):
A uniform solid cylinder of mass $M = 4.0\text{ kg}$ and radius $R = 0.30\text{ m}$ rests on a rough horizontal surface with coefficient of static friction $\mu_s = 0.30$. A horizontal pulling force $F = 12.0\text{ N}$ is applied at a height $h$ relative to the central axis ($-R \le h \le R$).
(a) Derive the general expressions for $a_{cm}$ and static friction $f_s$ assuming pure rolling.
(b) Determine the critical height $h_0$ at which no friction is required ($f_s = 0$).
(c) Calculate the magnitude and direction of $f_s$ for: (i) $h = R$ (top rim), (ii) $h = 0$ (center), and (iii) $h = -R/2$ (below center).
(d) For $h = R$, calculate the maximum pulling force $F_{\max}$ that can be applied without causing slipping.
Solution:
(a) General Formulation:
Assume static friction $f_s$ acts forward (in the $+x$ direction of $F$):
$$\sum F_x = F + f_s = M a_{cm} \quad \implies \quad f_s = M a_{cm} – F \quad \text{— (Equation 1)}$$
Torque about the centre of mass (taking clockwise as positive):
$$\tau_{cm} = F \cdot h – f_s \cdot R = I_{cm} \alpha = \left(\frac{1}{2} M R^2\right) \left(\frac{a_{cm}}{R}\right) = \frac{1}{2} M R a_{cm}$$
Divide by $R$:
$$F \left(\frac{h}{R}\right) – f_s = \frac{1}{2} M a_{cm} \quad \implies \quad f_s = F \left(\frac{h}{R}\right) – \frac{1}{2} M a_{cm} \quad \text{— (Equation 2)}$$
Equating Equation 1 and Equation 2:
$$M a_{cm} – F = F \left(\frac{h}{R}\right) – \frac{1}{2} M a_{cm} \implies \frac{3}{2} M a_{cm} = F \left(1 + \frac{h}{R}\right)$$
$$\mathbf{a_{cm} = \frac{2 F}{3 M} \left(1 + \frac{h}{R}\right)}$$
Substitute $a_{cm}$ into Equation 1 to find $f_s$:
$$f_s = M \left[\frac{2 F}{3 M} \left(1 + \frac{h}{R}\right)\right] – F = F \left[\frac{2}{3} + \frac{2h}{3R} – 1\right] = \mathbf{F \left(\frac{2h – R}{3R}\right)}$$
(b) Critical Height for Zero Friction ($f_s = 0$):
$$2h_0 – R = 0 \implies \mathbf{h_0 = \frac{R}{2} = 0.15\text{ m}}$$
When pulled at $R/2$ above the center, the torque $F(R/2)$ naturally produces the exact angular acceleration $\alpha = a_{cm}/R$ without needing any frictional assistance!
(c) Friction Evaluation for Specific Heights ($F = 12.0\text{ N}$, $M = 4.0\text{ kg}$):
– (i) $h = R$ (Top Rim):
$$f_s = 12.0 \left(\frac{2R – R}{3R}\right) = \frac{12.0}{3} = \mathbf{+4.0\text{ N}} \quad (\text{Forward, in direction of } F)$$
$$a_{cm} = \frac{2(12.0)}{3(4.0)}(1 + 1) = \mathbf{4.0\text{ m/s}^2}$$
Check: $F + f_s = 12.0 + 4.0 = 16.0\text{ N} = M a_{cm} = (4.0)(4.0) = 16.0\text{ N}$. (Verified!)
– (ii) $h = 0$ (Center of Cylinder):
$$f_s = 12.0 \left(\frac{0 – R}{3R}\right) = \mathbf{-4.0\text{ N}} \quad (\text{Backward, opposing } F)$$
$$a_{cm} = \frac{2(12.0)}{3(4.0)}(1 + 0) = \mathbf{2.0\text{ m/s}^2}$$
– (iii) $h = -R/2$ (Below Center):
$$f_s = 12.0 \left(\frac{2(-R/2) – R}{3R}\right) = 12.0 \left(\frac{-2R}{3R}\right) = \mathbf{-8.0\text{ N}} \quad (\text{Backward})$$
$$a_{cm} = \frac{2(12.0)}{3(4.0)}\left(1 – \frac{1}{2}\right) = \mathbf{1.0\text{ m/s}^2}$$
(d) Maximum Force for Pure Rolling at $h = R$:
Normal reaction: $N = M g = (4.0)(9.8) = 39.2\text{ N}$.
Maximum static friction available:
$$f_{s,\max} = \mu_s N = (0.30)(39.2) = 11.76\text{ N}$$
Since $f_s = F/3$:
$$\frac{F_{\max}}{3} = f_{s,\max} = 11.76 \implies \mathbf{F_{\max} = 3 \times 11.76 = 35.28\text{ N}}$$
Takeaway: Contrary to intuition, friction acts forward when a cylinder is pulled near the top, helping both translational acceleration and rotational balance.
Example 4 (Edge Case – Transition from Pure Sliding to Pure Rolling):
A uniform solid sphere of mass $M = 5.0\text{ kg}$ and radius $R = 0.20\text{ m}$ is projected horizontally on a rough horizontal floor with an initial linear velocity $v_0 = 14.0\text{ m/s}$ and zero initial angular velocity ($\omega_0 = 0$). The coefficient of kinetic friction is $\mu_k = 0.20$ (take $g = 9.8\text{ m/s}^2$).
(a) Determine the equations of motion for $v(t)$ and $\omega(t)$ during the slipping phase.
(b) Find the time $t_0$ when pure rolling begins and the common rolling speed $v_{\text{pure}}$.
(c) Calculate $v_{\text{pure}}$ using the conservation of angular momentum about an axis along the floor.
(d) Calculate the mechanical energy dissipated as heat during the slipping transition.
Solution:
(a) Slipping Dynamics:
Because $v_0 > 0$ and $\omega_0 = 0$, the contact point slips forward with speed $v_0$. Kinetic friction acts backward with constant magnitude:
$$f_k = \mu_k N = \mu_k M g = (0.20)(5.0)(9.8) = 9.80\text{ N}$$
Linear deceleration of the center of mass:
$$a = \frac{f_k}{M} = \mu_k g = (0.20)(9.8) = 1.96\text{ m/s}^2 \implies v(t) = v_0 – \mu_k g t$$
Angular acceleration about the $CM$:
$$\tau_{cm} = f_k R = I_{cm} \alpha = \left(\frac{2}{5} M R^2\right) \alpha \implies \alpha = \frac{5 \mu_k g}{2 R}$$
Since $\omega_0 = 0$:
$$\omega(t) = \alpha t = \left(\frac{5 \mu_k g}{2 R}\right) t$$
(b) Onset of Pure Rolling ($v(t_0) = R \omega(t_0)$):
$$v_0 – \mu_k g t_0 = R \left(\frac{5 \mu_k g}{2 R}\right) t_0 = \frac{5}{2} \mu_k g t_0$$
$$v_0 = \left(1 + \\frac{5}{2}\right) \mu_k g t_0 = \frac{7}{2} \mu_k g t_0$$
$$\mathbf{t_0 = \frac{2 v_0}{7 \mu_k g}} = \frac{2(14.0)}{7(0.20)(9.8)} = \frac{28.0}{13.72} \approx \mathbf{2.041\text{ s}}$$
The linear velocity when pure rolling is established:
$$v_{\text{pure}} = v_0 – \mu_k g \left(\frac{2 v_0}{7 \mu_k g}\right) = v_0 – \frac{2}{7} v_0 = \mathbf{\frac{5}{7} v_0} = \frac{5}{7}(14.0) = \mathbf{10.0\text{ m/s}}$$
(c) Shortcut via Conservation of Angular Momentum:
Consider a point $O$ lying on the horizontal floor. The normal force and gravity pass through the same vertical line, and friction acts along the floor through $O$. Thus, the net torque about point $O$ on the floor is zero ($\tau_O = 0$) throughout the entire motion! Hence, angular momentum about $O$ is strictly conserved:
$$L_{O, i} = L_{O, f}$$
Initially: $L_{O, i} = M v_0 R + I_{cm} \omega_0 = M v_0 R + 0 = M v_0 R$.
Finally (pure rolling): $L_{O, f} = M v_{\text{pure}} R + I_{cm} \omega_{\text{pure}} = M v_{\text{pure}} R + \left(\frac{2}{5} M R^2\right) \left(\frac{v_{\text{pure}}}{R}\right) = \frac{7}{5} M v_{\text{pure}} R$.
Equating both:
$$M v_0 R = \frac{7}{5} M v_{\text{pure}} R \implies \mathbf{v_{\text{pure}} = \frac{5}{7} v_0 = 10.0\text{ m/s}}$$
(Notice how this elegant one-line calculation does not depend on $\mu_k$ or $g$!).
(d) Energy Dissipation:
Initial mechanical energy:
$$K_i = \frac{1}{2} M v_0^2 = \frac{1}{2} (5.0\text{ kg})(14.0\text{ m/s})^2 = \frac{1}{2} (5.0)(196.0) = \mathbf{490.0\text{ Joules}}$$
Final mechanical energy:
$$K_f = \frac{1}{2} M v_{\text{pure}}^2 \left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2} (5.0)(10.0)^2 \left(1 + \frac{2}{5}\right) = (250.0)(1.4) = \mathbf{350.0\text{ Joules}}$$
Mechanical energy dissipated as heat:
$$\Delta K = K_i – K_f = 490.0 – 350.0 = \mathbf{140.0\text{ Joules}}$$
Takeaway: Choosing an origin along the contact plane eliminates the torque of friction, enabling instantaneous determination of final rolling speed via angular momentum conservation.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A uniform circular disc of mass $M$ and radius $R$ rolls without slipping on a horizontal surface with linear speed $v$. The ratio of its rotational kinetic energy about its centre of mass to its total kinetic energy is:
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{3}$
(D) $\frac{1}{4}$
Problem 2 (JEE Main – Single Correct):
A sphere of radius $R$ is rolling without slipping on a horizontal plane with center-of-mass velocity $v$. The velocity of the lowest point of contact on the sphere with the plane is:
(A) $v$ forward
(B) $2v$ forward
(C) Zero
(D) $v$ backward
Problem 3 (JEE Main – Single Correct):
A uniform solid sphere of mass $M$ and radius $R$ rolls down an inclined plane of inclination $\theta$ without slipping. The linear acceleration of its centre of mass is:
(A) $\frac{5}{7} g \sin\theta$
(B) $\frac{2}{3} g \sin\theta$
(C) $\frac{1}{2} g \sin\theta$
(D) $\frac{5}{9} g \sin\theta$
Problem 4 (JEE Main – Single Correct):
A thin circular ring and a uniform circular disc of equal mass and equal radius are released from rest from the same height on a rough inclined plane and roll down without slipping. The ratio of their velocities at the bottom of the incline ($v_{\text{ring}} / v_{\text{disc}}$) is:
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{2}{\sqrt{3}}$
(C) $\sqrt{\frac{2}{3}}$
(D) $\frac{3}{4}$
Problem 5 (JEE Main – Single Correct):
A uniform cylinder of radius $R$ rolls without slipping on a horizontal ground with speed $v$. The speed of the highest point on the cylinder is:
(A) $v$
(B) $\sqrt{2} v$
(C) $2 v$
(D) $4 v$
Problem 6 (JEE Advanced – One or More Correct):
A rigid symmetric body rolls without slipping on a stationary horizontal ground. Which of the following statements is/are correct?
(A) The instantaneous linear velocity of the contact point is zero.
(B) The instantaneous linear acceleration of the contact point is non-zero and directed toward the centre of the body ($a_c = \omega^2 R$).
(C) The total work done by static friction on the body over any finite time interval is zero.
(D) The total kinetic energy can be expressed as $\frac{1}{2} I_C \omega^2$, where $I_C$ is the moment of inertia about the instantaneous axis passing through contact point $C$.
Problem 7 (JEE Advanced – One or More Correct):
Four bodies of identical mass $M$ and radius $R$—a solid sphere, a solid cylinder, a thin spherical shell, and a thin circular ring—roll down a rough inclined plane of inclination $\theta$ without slipping from rest:
(A) The solid sphere reaches the bottom first.
(B) The thin circular ring reaches the bottom last.
(C) The magnitude of the static friction force required is greatest for the thin circular ring.
(D) At the bottom of the incline, all four bodies have identical total mechanical kinetic energy ($M g h$).
Problem 8 (JEE Advanced – One or More Correct):
A spool of mass $M$, inner hub radius $r$, and outer rim radius $R$ rests on a rough horizontal floor. A light string wound around the inner hub is pulled with a tension force $F$ at an angle $\theta$ to the horizontal ($0 \le \theta \le 90^\circ$):
(A) If the line of action of the pulling force passes above the contact point $C$, the spool rolls forward.
(B) If the line of action of the pulling force passes below the contact point $C$, the spool rolls backward.
(C) The critical angle $\theta_0$ at which the spool has no tendency to roll satisfies $\cos\theta_0 = \frac{r}{R}$.
(D) At $\cos\theta_0 = \frac{r}{R}$, the spool is in limiting static equilibrium or skids without rolling.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform solid sphere of mass $M = 7.0\text{ kg}$ rolls without slipping on a horizontal floor with a constant linear speed $v = 10.0\text{ m/s}$. Calculate the total kinetic energy of the sphere in Joules.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid cylinder of mass $M = 3.0\text{ kg}$ and radius $R = 0.40\text{ m}$ is placed on a rough inclined plane of inclination $\theta = 30^\circ$. The cylinder rolls down the incline without slipping. Taking $g = 10.0\text{ m/s}^2$, calculate the magnitude of the static friction force $f_s$ (in Newtons) acting on the cylinder.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (A) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 490 | 10. 5
Solution 1:
For a disc, $I_{cm} = \frac{1}{2} M R^2$.
$$K_{\text{rot}} = \frac{1}{2} I_{cm} \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4} M v^2$$
$$K_{\text{trans}} = \frac{1}{2} M v^2$$
$$K_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$$
$$\text{Ratio} = \frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{4} M v^2}{\frac{3}{4} M v^2} = \mathbf{\frac{1}{3}}$$
Correct Option: (B)
Solution 2:
In pure rolling, the contact point $C$ has velocity $v_C = v_{cm} – \omega R$. Since $v_{cm} = \omega R$, $v_C = 0$. The contact point is instantaneously at rest.
Correct Option: (C)
Solution 3:
For pure rolling down an incline:
$$a_{cm} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}$$
For a solid sphere, $\frac{k^2}{R^2} = \frac{2}{5}$:
$$a_{cm} = \frac{g \sin\theta}{1 + \frac{2}{5}} = \mathbf{\frac{5}{7} g \sin\theta}$$
Correct Option: (A)
Solution 4:
By conservation of mechanical energy, $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$.
For ring: $k^2/R^2 = 1 \implies v_{\text{ring}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}$.
For disc: $k^2/R^2 = 1/2 \implies v_{\text{disc}} = \sqrt{\frac{2gh}{1 + 1/2}} = \sqrt{\frac{4gh}{3}}$.
$$\frac{v_{\text{ring}}}{v_{\text{disc}}} = \frac{\sqrt{gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{\frac{3}{4}} = \mathbf{\frac{\sqrt{3}}{2}}$$
Correct Option: (A)
Solution 5:
At the topmost point, $v_{\text{top}} = v_{cm} + \omega R = v + v = \mathbf{2v}$.
Alternatively, via IAOR at the bottom, distance is $2R$: $v_{\text{top}} = \omega (2R) = 2(\omega R) = 2v$.
Correct Option: (C)
Solution 6:
– (A) True: Pure rolling requires zero relative velocity at contact: $v_C = 0$.
– (B) True: Even though $v_C = 0$, the contact point moves along a cycloid. Its acceleration is pure centripetal acceleration directed toward the center: $a_c = \omega^2 R$.
– (C) True: $dW = \vec{f}_s \cdot d\vec{s}_C = \vec{f}_s \cdot (\vec{v}_C dt) = 0$ since $v_C = 0$.
– (D) True: $K = \frac{1}{2} I_C \omega^2 = \frac{1}{2}(I_{cm} + M R^2)\omega^2 = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$.
All four statements are correct.
Correct Options: (A, B, C, D)
Solution 7:
– (A, B) True: $a_{cm} = \frac{g\sin\theta}{1 + k^2/R^2}$. Values of $k^2/R^2$: Solid sphere ($0.40$), Solid cylinder ($0.50$), Spherical shell ($0.67$), Ring ($1.00$). The sphere has the highest acceleration and arrives first; the ring has the lowest acceleration and arrives last.
– (C) True: $f_s = M g \sin\theta / (1 + R^2/k^2) = M g \sin\theta \frac{k^2/R^2}{1 + k^2/R^2}$. For the ring ($k^2/R^2 = 1$), $f_s = \frac{1}{2} M g \sin\theta$, which is the highest friction force among the four bodies.
– (D) True: Because static friction does zero work, mechanical energy is conserved for all four bodies: $K_{\text{total}} = M g h$.
All four statements are correct.
Correct Options: (A, B, C, D)
Solution 8:
Taking torque about the contact point $C$ on the floor:
The moment arm of $F$ about $C$ is $d = R \cos\theta – r$.
– If $\cos\theta > r/R$, the torque about $C$ is clockwise, so the spool rolls forward (A is correct).
– If $\cos\theta < r/R$, the torque about $C$ is counterclockwise, so the spool rolls backward (B is correct).
– If $\cos\theta_0 = r/R$, the line of action of $F$ passes directly through $C$, producing zero torque about $C$ ($d = 0$), so the spool cannot roll (C, D are correct).
All four options are correct.
Correct Options: (A, B, C, D)
Solution 9:
For a uniform solid sphere, $k^2/R^2 = 2/5$.
Total kinetic energy:
$$K = \frac{1}{2} M v^2 \left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2} (7.0\text{ kg})(10.0\text{ m/s})^2 \left(1 + \frac{2}{5}\right)$$
$$K = \frac{1}{2} (7.0)(100.0)(1.4) = (350.0)(1.4) = \mathbf{490\text{ Joules}}$$
Correct Answer: 490
Solution 10:
For a solid cylinder rolling down an incline of angle $\theta = 30^\circ$ ($k^2/R^2 = 1/2$):
$$f_s = \frac{M g \sin\theta}{1 + \frac{R^2}{k^2}} = \frac{M g \sin 30^\circ}{1 + 2} = \frac{1}{3} M g \sin 30^\circ$$
Substitute values ($M = 3.0\text{ kg}$, $g = 10.0\text{ m/s}^2$, $\sin 30^\circ = 0.5$):
$$f_s = \frac{1}{3} (3.0)(10.0)(0.5) = \frac{1}{3} (15.0) = \mathbf{5.0\text{ N}}$$
Correct Answer: 5