Concept Card: Combined Mechanics in Rotational Motion – Advanced Problem-Solving Framework
1. The Synthesis Framework for Complex Rotational Mechanics:
Rotational motion problems in JEE Advanced rarely appear in isolation; they integrate translational dynamics, torque equations, impulse-momentum theorems, work-energy principles, and non-inertial frames. A structured 5-step analytical algorithm guarantees consistent success across all combined mechanics problems:
- Step 1: Free Body Diagram (FBD) with Geometric Points of Application:
Unlike particle mechanics where forces can be treated as concurrent at a point mass, rigid body mechanics requires drawing each force at its exact point of application. Contact forces (normal reactions, friction, string tensions) act at physical interfaces, while gravity acts through the Center of Gravity ($CG$). - Step 2: Linear Kinematic & Dynamic Equations of the Center of Mass:
Apply Newton’s Second Law for the translation of the system:
$$\sum \vec{F}_{\text{ext}} = M \vec{a}_{cm} \implies \sum F_x = M a_{cm, x}, \quad \sum F_y = M a_{cm, y}$$ - Step 3: Rotational Dynamic Equations (Choice of Reference Point):
Apply the rotational equation of motion:
$$\sum \vec{\tau}_P = I_P \vec{\alpha} + \vec{r}_{cm/P} \times (-M \vec{a}_P)$$
Crucial Theorem on Pivot Selection: The pseudo-torque term $\vec{r}_{cm/P} \times (-M \vec{a}_P)$ vanishes, allowing the simplified form $\sum \vec{\tau}_P = I_P \vec{\alpha}$, if and only if reference point $P$ is:- An inertial fixed point in space ($\vec{a}_P = \vec{0}$, such as a fixed hinge).
- The Center of Mass of the body ($\vec{r}_{cm/P} = \vec{0}$, valid in all frames, even accelerating ones!).
- A point whose acceleration $\vec{a}_P$ is directed straight toward or away from the $CM$.
- Step 4: Kinematic Constraint Equations:
Relate translational acceleration to angular acceleration using contact and geometry constraints:- Unwinding string from a drum/pulley: $a = \alpha R$.
- Pure rolling on stationary ground: $a_{cm} = \alpha R$ and $v_{cm} = \omega R$.
- Rolling on a moving plank / surface: $\vec{a}_{\text{contact}} = \vec{a}_{cm} + \vec{\alpha} \times \vec{R} = \vec{a}_{\text{surface}}$.
- Step 5: Work-Energy & Conservation Principles:
When internal non-conservative forces do no work:
$$E_{\text{mech}} = K_{\text{trans}} + K_{\text{rot}} + U = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2 + M g y_{cm} = \text{constant}$$
Remember that static friction in pure rolling on stationary surfaces does zero work ($W_{f_s} = 0$).
2. Total Angular Momentum of an Arbitrary Moving Body
For an extended rigid body executing combined translation and rotation in plane motion, its total angular momentum about an arbitrary reference origin $O$ is resolved into two distinct vector contributions:
Chasles-Koenig Angular Momentum Theorem:
$$\mathbf{\vec{L}_O = \vec{L}_{\text{orbital}} + \vec{L}_{\text{spin}} = (\vec{r}_{cm} \times M \vec{v}_{cm}) + I_{cm} \vec{\omega}}$$
where:
- $\vec{L}_{\text{orbital}} = \vec{r}_{cm} \times M \vec{v}_{cm}$ is the angular momentum of the center of mass treated as a point particle carrying the entire mass $M$.
- $\vec{L}_{\text{spin}} = I_{cm} \vec{\omega}$ is the intrinsic spin angular momentum of the body rotating about its center of mass.
Conservation Criterion: $\vec{L}_O$ is conserved if and only if the net external torque about that specific point $O$ is zero ($\sum \vec{\tau}_{O, \text{ext}} = \vec{0}$).
3. Impulsive Collisions and Ballistic Impact Dynamics
When high-speed impulsive collisions occur with extended rigid bodies:
- Free Bodies (Unconstrained): Both linear momentum $\vec{p}$ and angular momentum about the $CM$ ($\vec{L}_{cm}$) are conserved.
- Hinged Bodies (Constrained Pivots):
- Linear momentum is NOT conserved because the hinge delivers an instantaneous external impulsive reaction force $\vec{J}_{\text{hinge}}$.
- Angular momentum about the hinge pivot axis $O$ IS strictly conserved because the line of action of the hinge impulse passes directly through the pivot ($r_{\text{hinge}} = 0 \implies \tau_{\text{hinge}} = 0$):
$$\mathbf{L_{O, \text{initial}} = L_{O, \text{final}}}$$
- Center of Percussion (Sweet Spot):
For any hinged rigid body, there exists a unique point along its length known as the Center of Percussion ($COP$) such that an impact delivered at this point produces zero impulsive reaction force at the hinge ($J_{\text{hinge}} = 0$).
For a body pivoted at $O$ with center of mass at distance $d_{cm}$ from $O$:
$$\mathbf{x_{cop} = \frac{I_O}{M d_{cm}} = \frac{k_O^2}{d_{cm}}}$$
For a uniform thin rod of length $L$ pivoted at one end ($I_O = \frac{1}{3} M L^2$, $d_{cm} = L/2$):
$$x_{cop} = \frac{\frac{1}{3} M L^2}{M (L/2)} = \mathbf{\frac{2}{3} L}$$
4. Essential Insights & Pitfalls for JEE Advanced
- Trap 1 (Using Torque about an Accelerating Point): Never write $\tau_P = I_P \alpha$ about an arbitrary point $P$ unless $P$ is fixed in an inertial frame or $P$ is the $CM$. If you must choose an accelerating pivot $P$, you must apply an inertial pseudo-force $-M \vec{a}_P$ at the $CM$.
- Trap 2 (Misidentifying the Direction of Static Friction): In combined rolling problems, static friction adjusts in magnitude and direction to satisfy $a_{cm} = \alpha R$. Do not assume friction opposes the pulling force.
- Trap 3 (Energy Loss in Inelastic Rotational Collisions): In ballistic collisions where a projectile embeds itself into a rigid body, mechanical energy is never conserved. Kinetic energy is dissipated into internal heat and plastic deformation; only angular momentum about the pivot is conserved.
Solved Examples
Example 1 (Direct Multi-Concept Application – Stepped Real Pulley with Suspended Masses):
A stepped compound pulley consists of two concentric cylinders rigidly joined together, mounted on a frictionless horizontal axle. The outer drum has radius $R_1 = 0.20\text{ m}$, and the inner drum has radius $R_2 = 0.10\text{ m}$. The total moment of inertia of the compound pulley about its central axle is $I = 0.16\text{ kg}\cdot\text{m}^2$. A block of mass $m_1 = 3.0\text{ kg}$ hangs from a light cord wrapped around the outer drum, while a block of mass $m_2 = 2.0\text{ kg}$ hangs from a cord wrapped around the inner drum in the opposing sense (so that as $m_1$ descends, $m_2$ ascends). (Take $g = 9.8\text{ m/s}^2$).
(a) Determine the angular acceleration $\alpha$ of the pulley and the linear accelerations $a_1$ and $a_2$ of the two blocks.
(b) Calculate the cord tensions $T_1$ and $T_2$.
(c) If the system is released from rest, calculate the speed $v_1$ of mass $m_1$ after it descends through a vertical distance $h_1 = 2.0\text{ m}$ using the principle of conservation of mechanical energy.
Solution:
(a) Equations of Motion:
Let $\alpha$ be the clockwise angular acceleration of the pulley.
– For block $m_1$ (unwinding from outer drum, descending with $a_1 = \alpha R_1$):
$$m_1 g – T_1 = m_1 a_1 = m_1 (\alpha R_1) \implies T_1 = m_1 (g – \alpha R_1) \quad \text{— (Equation 1)}$$
– For block $m_2$ (winding onto inner drum, ascending with $a_2 = \alpha R_2$):
$$T_2 – m_2 g = m_2 a_2 = m_2 (\alpha R_2) \implies T_2 = m_2 (g + \alpha R_2) \quad \text{— (Equation 2)}$$
– For the compound pulley (rotational dynamics about fixed central axle):
$$\tau_{\text{net}} = T_1 R_1 – T_2 R_2 = I \alpha \quad \text{— (Equation 3)}$$
Substitute Equations 1 and 2 into Equation 3:
$$[m_1 (g – \alpha R_1)] R_1 – [m_2 (g + \alpha R_2)] R_2 = I \alpha$$
$$m_1 g R_1 – m_1 R_1^2 \alpha – m_2 g R_2 – m_2 R_2^2 \alpha = I \alpha$$
$$g (m_1 R_1 – m_2 R_2) = (I + m_1 R_1^2 + m_2 R_2^2) \alpha$$
$$\mathbf{\alpha = \frac{g (m_1 R_1 – m_2 R_2)}{I + m_1 R_1^2 + m_2 R_2^2}}$$
Substitute numerical values ($m_1 = 3.0\text{ kg}, R_1 = 0.20\text{ m}, m_2 = 2.0\text{ kg}, R_2 = 0.10\text{ m}, I = 0.16\text{ kg}\cdot\text{m}^2$):
$$\text{Driving Torque} = 9.8 [(3.0)(0.20) – (2.0)(0.10)] = 9.8 [0.60 – 0.20] = 9.8(0.40) = 3.92\text{ N}\cdot\text{m}$$
$$\text{Effective Inertia} = 0.16 + (3.0)(0.20)^2 + (2.0)(0.10)^2 = 0.16 + 0.12 + 0.02 = 0.30\text{ kg}\cdot\text{m}^2$$
$$\alpha = \frac{3.92\text{ N}\cdot\text{m}}{0.30\text{ kg}\cdot\text{m}^2} = \mathbf{13.07\text{ rad/s}^2}$$
Linear accelerations:
$$a_1 = \alpha R_1 = (13.067)(0.20) = \mathbf{2.61\text{ m/s}^2}$$
$$a_2 = \alpha R_2 = (13.067)(0.10) = \mathbf{1.31\text{ m/s}^2}$$
(b) Cord Tensions:
$$T_1 = m_1 (g – a_1) = 3.0(9.8 – 2.613) = 3.0(7.187) = \mathbf{21.56\text{ N}}$$
$$T_2 = m_2 (g + a_2) = 2.0(9.8 + 1.307) = 2.0(11.107) = \mathbf{22.21\text{ N}}$$
(c) Speed $v_1$ via Conservation of Mechanical Energy:
When $m_1$ descends by $h_1 = 2.0\text{ m}$, the angle turned is $\theta = \frac{h_1}{R_1} = \frac{2.0}{0.20} = 10.0\text{ rad}$.
Block $m_2$ rises by $h_2 = \theta R_2 = (10.0)(0.10) = 1.0\text{ m}$.
Using kinematic formula for constant acceleration:
$$v_1 = \sqrt{2 a_1 h_1} = \sqrt{2(2.6133)(2.0)} = \sqrt{10.453} = \mathbf{3.23\text{ m/s}}$$
(Energy verification: Loss in PE $= m_1 g h_1 – m_2 g h_2 = (3.0)(9.8)(2.0) – (2.0)(9.8)(1.0) = 58.8 – 19.6 = 39.2\text{ J}$. Gain in KE $= \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{1}{2} I \omega^2 = \frac{1}{2}(0.30)\omega^2 = \frac{1}{2}(0.30)(v_1/R_1)^2 = \frac{1}{2}(0.30)(10.453 / 0.04) = 39.2\text{ J}$. Confirmed!).
Takeaway: In complex pulley systems, write the effective rotational inertia of the system as $I_{\text{eff}} = I + \sum m_i R_i^2$, instantly yielding the system’s angular acceleration.
Example 2 (Mathematical Formulation – Inelastic Ballistic Collision with Free Rigid Rod):
A uniform thin rod of mass $M = 2.0\text{ kg}$ and length $L = 1.20\text{ m}$ lies completely at rest on a frictionless horizontal ice rink. A small sticky projectile of mass $m = 0.50\text{ kg}$ sliding with horizontal speed $v_0 = 10.0\text{ m/s}$ perpendicular to the rod strikes one of the ends of the rod and remains embedded in it.
(a) Determine the position of the new Center of Mass ($CM’$) of the combined system.
(b) Calculate the linear velocity $v_{cm}’$ of the combined system after the impact.
(c) Calculate the moment of inertia $I_{cm}’$ of the combined system about its new center of mass.
(d) Find the post-collision angular velocity $\omega’$ of the combined system.
(e) Determine the loss of mechanical kinetic energy during the collision.
Solution:
(a) Location of New Center of Mass:
Let the initial center of the rod be the origin ($y = 0$). The rod’s mass $M = 2.0\text{ kg}$ is at $y = 0$, and the projectile $m = 0.50\text{ kg}$ embeds at end $y = +L/2 = +0.60\text{ m}$.
$$y_{cm}’ = \frac{M(0) + m(L/2)}{M + m} = \frac{0 + (0.50)(0.60)}{2.0 + 0.50} = \frac{0.30}{2.50} = \mathbf{0.12\text{ m}}$$
The new center of mass lies along the rod at a distance of $0.12\text{ m}$ from its geometric center toward the struck end.
(b) Velocity of Center of Mass $v_{cm}’$:
Because no external horizontal forces act on the system during the impact, linear momentum is conserved:
$$m v_0 = (M + m) v_{cm}’ \implies v_{cm}’ = \frac{m v_0}{M + m} = \frac{(0.50)(10.0)}{2.50} = \mathbf{2.00\text{ m/s}}$$
(c) Moment of Inertia $I_{cm}’$ of Combined Body About New $CM’$:
– Rod about new $CM’$ (using Parallel Axis Theorem, shift $d = 0.12\text{ m}$):
$$I_{\text{rod}} = \frac{1}{12} M L^2 + M (0.12)^2 = \frac{1}{12}(2.0)(1.20)^2 + 2.0(0.0144) = 0.240 + 0.0288 = 0.2688\text{ kg}\cdot\text{m}^2$$
– Embedded projectile about new $CM’$ (distance $r = 0.60 – 0.12 = 0.48\text{ m}$):
$$I_{\text{particle}} = m r^2 = (0.50)(0.48)^2 = (0.50)(0.2304) = 0.1152\text{ kg}\cdot\text{m}^2$$
Total moment of inertia:
$$I_{cm}’ = 0.2688 + 0.1152 = \mathbf{0.384\text{ kg}\cdot\text{m}^2}$$
(d) Post-Collision Angular Velocity $\omega’$:
Conserve angular momentum about the system’s new center of mass $CM’$ (which is an unaccelerated reference point before the collision, and about which external torque is zero):
Before collision: the projectile moves with velocity $v_0$ at perpendicular distance $r = 0.48\text{ m}$ from $CM’$:
$$L_{cm’, \text{initial}} = m v_0 r = (0.50)(10.0)(0.48) = 2.40\text{ kg}\cdot\text{m}^2/\text{s}$$
After collision:
$$L_{cm’, \text{final}} = I_{cm}’ \omega’$$
$$2.40 = (0.384) \omega’ \implies \mathbf{\omega’ = \frac{2.40}{0.384} = 6.25\text{ rad/s}}$$
(e) Loss of Kinetic Energy:
– Initial kinetic energy:
$$K_{\text{initial}} = \frac{1}{2} m v_0^2 = \frac{1}{2}(0.50)(10.0)^2 = 25.0\text{ J}$$
– Final kinetic energy:
$$K_{\text{final}} = \frac{1}{2}(M + m)(v_{cm}’)^2 + \frac{1}{2} I_{cm}’ (\omega’)^2$$
$$K_{\text{trans}} = \frac{1}{2}(2.50)(2.0)^2 = 5.0\text{ J}$$
$$K_{\text{rot}} = \frac{1}{2}(0.384)(6.25)^2 = \frac{1}{2}(0.384)(39.0625) = 7.5\text{ J}$$
$$K_{\text{final}} = 5.0 + 7.5 = 12.5\text{ J}$$
$$\mathbf{\Delta K = K_{\text{initial}} – K_{\text{final}} = 25.0 – 12.5 = 12.5\text{ J}}$$
Exactly $50\%$ of the initial mechanical energy is converted to internal energy.
Example 3 (Advanced JEE Analysis – Hinged Ballistic Impact & Center of Percussion):
A uniform slender rod of mass $M = 3.0\text{ kg}$ and length $L = 1.0\text{ m}$ is pivoted smoothly at its upper end $O$ so that it hangs vertically at rest. A horizontal bullet of mass $m = 0.10\text{ kg}$ travelling at $v_0 = 100.0\text{ m/s}$ strikes the rod at a distance $x$ below the hinge $O$ and emerges horizontally with a reduced speed of $v_1 = 20.0\text{ m/s}$.
(a) Derive the expression for the angular velocity $\omega$ of the rod immediately after the impact.
(b) Derive the expression for the horizontal impulsive reaction force $J_H$ exerted by the hinge on the rod during the impact.
(c) Find the specific impact distance $x = x_{cop}$ for which the horizontal hinge impulse $J_H$ is identically zero (the Center of Percussion).
(d) If $x = x_{cop}$, calculate $\omega$ and the maximum angle of deflection $\theta_{\text{max}}$ of the rod.
Solution:
(a) Angular Velocity via Conservation of Angular Momentum about Hinge $O$:
The hinge exerts an unknown external impulse $\vec{J}_H$, but its torque about pivot $O$ is zero ($\tau_O = 0$). Hence, angular momentum about $O$ is conserved:
$$L_{O, \text{initial}} = L_{O, \text{final}}$$
$$m v_0 x = m v_1 x + I_O \omega$$
where $I_O = \frac{1}{3} M L^2$.
$$m(v_0 – v_1) x = \frac{1}{3} M L^2 \omega \implies \mathbf{\omega = \frac{3 m (v_0 – v_1) x}{M L^2}}$$
(b) Horizontal Hinge Impulse $J_H$:
Let $J_H$ be directed in the direction of bullet motion ($+x$).
The linear impulse-momentum theorem applied to the rod in the horizontal direction gives:
$$J_{\text{bullet on rod}} + J_H = M v_{cm, \text{final}}$$
The impulse delivered by the bullet onto the rod is $J_b = m(v_0 – v_1)$.
The rod’s center of mass is at $d_{cm} = L/2$, so its linear velocity immediately after impact is:
$$v_{cm} = \omega d_{cm} = \omega \left(\frac{L}{2}\right)$$
Substitute $\omega$ from part (a):
$$v_{cm} = \left[\frac{3 m(v_0 – v_1) x}{M L^2}\right] \left(\frac{L}{2}\right) = \frac{3 m(v_0 – v_1) x}{2 M L}$$
Hence:
$$m(v_0 – v_1) + J_H = M \left[\frac{3 m(v_0 – v_1) x}{2 M L}\right] = \frac{3 m(v_0 – v_1) x}{2 L}$$
$$\mathbf{J_H = m(v_0 – v_1) \left(\frac{3x}{2L} – 1\right)}$$
(c) Condition for Zero Hinge Impulse (Center of Percussion):
Setting $J_H = 0$:
$$\frac{3x}{2L} – 1 = 0 \implies \mathbf{x_{cop} = \frac{2}{3} L}$$
For $L = 1.0\text{ m}$:
$$x_{cop} = \frac{2}{3}(1.0\text{ m}) = \mathbf{0.667\text{ m}}$$
When struck at two-thirds of its length, the rod naturally rotates around the hinge without trying to shove the hinge forward or pull it backward!
(d) Numerical Calculations for $x = 2/3\text{ m}$:
Impulse delivered by bullet:
$$J_b = (0.10\text{ kg})(100.0 – 20.0\text{ m/s}) = 8.0\text{ N}\cdot\text{s}$$
Angular velocity $\omega$:
$$\omega = \frac{3(8.0)(2/3)}{(3.0)(1.0)^2} = \frac{16.0}{3.0} = \mathbf{5.33\text{ rad/s}}$$
Maximum swing angle $\theta_{\text{max}}$ via conservation of energy during post-collision swing:
$$\frac{1}{2} I_O \omega^2 = M g \left(\frac{L}{2}\right) (1 – \cos\theta_{\text{max}})$$
$$\frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2 = M g \frac{L}{2} (1 – \cos\theta_{\text{max}})$$
$$\frac{1}{6} L \omega^2 = \frac{1}{2} g (1 – \cos\theta_{\text{max}}) \implies 1 – \cos\theta_{\text{max}} = \frac{L \omega^2}{3g}$$
$$1 – \cos\theta_{\text{max}} = \frac{(1.0)(5.333)^2}{3(9.8)} = \frac{28.44}{29.4} = 0.9675$$
$$\cos\theta_{\text{max}} = 1 – 0.9675 = 0.0325 \implies \mathbf{\theta_{\text{max}} \approx 88.1^\circ}$$
Example 4 (Complex Coupled Dynamics – Cylinder Rolling on an Accelerating Plank):
A plank of mass $m_p = 4.0\text{ kg}$ rests on a frictionless horizontal floor. A uniform solid cylinder of mass $M = 6.0\text{ kg}$ and radius $R = 0.25\text{ m}$ rests on top of the plank. The coefficient of static friction between the cylinder and the plank is $\mu_s = 0.50$. A constant horizontal force $F = 40.0\text{ N}$ is applied to the plank.
(a) Assuming no slipping occurs between the cylinder and the plank, derive expressions for the acceleration of the plank $a_p$, the acceleration of the center of mass of the cylinder $a_c$, the angular acceleration $\alpha$ of the cylinder, and the friction force $f_s$.
(b) Calculate the numerical values for $a_p$, $a_c$, $\alpha$, and $f_s$.
(c) Determine the maximum pulling force $F_{\text{max}}$ that can be applied before the cylinder begins to slip on the plank.
Solution:
(a) Equations of Motion and Constraint:
Let $F$ pull the plank to the right ($+x$).
– Action-Reaction at Contact Surface: By Newton’s Third Law, if friction $f_s$ acts to the right on the cylinder, an equal and opposite friction force $f_s$ acts to the left on the plank.
– For the cylinder:
1. Translational motion of $CM$:
$$f_s = M a_c \quad \text{— (Equation 1)}$$
2. Rotational motion about $CM$ (torque from $f_s$ opposes forward roll, causing counter-clockwise angular acceleration $\alpha$):
$$\tau_{cm} = f_s R = I_{cm} \alpha = \left(\frac{1}{2} M R^2\right) \alpha \implies f_s = \frac{1}{2} M (\alpha R) \quad \text{— (Equation 2)}$$
Comparing Equations 1 and 2:
$$M a_c = \frac{1}{2} M (\alpha R) \implies \mathbf{\alpha R = 2 a_c}$$
– Kinematic Rolling Constraint (No Slipping):
The bottom point of the cylinder is moving forward with velocity of $CM$ minus tangential speed of roll, so its forward acceleration is:
$$a_{\text{bottom}} = a_c – \alpha R = a_c – 2a_c = -a_c$$
For no slipping between the cylinder contact point and the plank:
$$a_{\text{bottom}} = a_{\text{plank}} \implies a_p = a_c – (-\alpha R) = a_c + \alpha R$$
Using $\alpha R = 2a_c$:
$$\mathbf{a_p = a_c + 2a_c = 3a_c \implies a_c = \frac{1}{3} a_p}$$
– For the plank:
$$F – f_s = m_p a_p \quad \text{— (Equation 3)}$$
Substitute $f_s = M a_c = M \left(\frac{1}{3} a_p\right) = \frac{1}{3} M a_p$ into Equation 3:
$$F – \frac{1}{3} M a_p = m_p a_p \implies F = \left(m_p + \frac{1}{3} M\right) a_p$$
$$\mathbf{a_p = \frac{F}{m_p + \frac{1}{3} M}}$$
$$\mathbf{a_c = \frac{F}{3 m_p + M}}$$
$$\mathbf{f_s = M a_c = \frac{M F}{3 m_p + M}}$$
(b) Numerical Evaluations:
With $m_p = 4.0\text{ kg}, M = 6.0\text{ kg}, F = 40.0\text{ N}, R = 0.25\text{ m}$:
$$a_p = \frac{40.0}{4.0 + \frac{1}{3}(6.0)} = \frac{40.0}{4.0 + 2.0} = \frac{40.0}{6.0} = \mathbf{6.67\text{ m/s}^2}$$
$$a_c = \frac{1}{3} a_p = \frac{6.67}{3} = \mathbf{2.22\text{ m/s}^2}$$
$$\alpha = \frac{2 a_c}{R} = \frac{2(2.222)}{0.25} = \mathbf{17.78\text{ rad/s}^2}$$
$$f_s = M a_c = (6.0)(2.222) = \mathbf{13.33\text{ N}}$$
(c) Maximum Applied Force $F_{\text{max}}$ Without Slipping:
The maximum available static friction is:
$$f_{s, \text{max}} = \mu_s N = \mu_s M g = (0.50)(6.0)(9.8) = 29.4\text{ N}$$
Setting $f_s \le f_{s, \text{max}}$:
$$\frac{M F_{\text{max}}}{3 m_p + M} = f_{s, \text{max}}$$
$$F_{\text{max}} = f_{s, \text{max}} \left(\frac{3 m_p + M}{M}\right) = 29.4 \left(\frac{3(4.0) + 6.0}{6.0}\right) = 29.4 \left(\frac{18.0}{6.0}\right) = 29.4(3) = \mathbf{88.2\text{ N}}$$
Practice Worksheet: Combined Mechanics & Advanced Rotational Dynamics
Section A: Single Correct Questions (Q1 to Q5)
Q1. A uniform circular disc of mass $M$ and radius $R$ is placed on a smooth horizontal table. A horizontal force $F$ is applied tangentially at the highest point of the rim. What is the angular speed of the disc after time $t$?
(A) $\frac{Ft}{MR}$
(B) $\frac{2Ft}{MR}$
(C) $\frac{4Ft}{MR}$
(D) $\frac{Ft}{2MR}$
Q2. A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping up a rough inclined plane of inclination angle $\theta$. If its initial linear speed at the bottom is $v$, the maximum vertical height $h$ attained by the center of mass of the sphere before coming momentarily to rest is:
(A) $\frac{v^2}{2g}$
(B) $\frac{v^2}{g}$
(C) $\frac{7v^2}{10g}$
(D) $\frac{5v^2}{7g}$
Q3. A uniform thin rod of mass $M$ and length $L$ lies at rest on a smooth horizontal ice surface. A horizontal impulse $J$ is applied perpendicular to the rod at a distance $L/4$ from its center. The resulting angular velocity $\omega$ of the rod immediately after the impulse is:
(A) $\frac{J}{ML}$
(B) $\frac{2J}{ML}$
(C) $\frac{3J}{ML}$
(D) $\frac{6J}{ML}$
Q4. A particle of mass $m$ is projected with speed $u$ at an angle $\theta$ above the horizontal. The magnitude of the angular momentum of the projectile about the launch point when it reaches its highest point is:
(A) $\frac{m u^3 \sin^2\theta \cos\theta}{2g}$
(B) $\frac{m u^3 \sin\theta \cos^2\theta}{2g}$
(C) $\frac{m u^3 \sin^2\theta \cos\theta}{g}$
(D) $\frac{m u^3 \sin\theta \cos\theta}{2g}$
Q5. A thin ring of radius $R$ rolls without slipping down an inclined plane of inclination $\theta$. The minimum coefficient of static friction $\mu_{\text{min}}$ between the ring and the incline required to sustain pure rolling is:
(A) $\frac{1}{3}\tan\theta$
(B) $\frac{1}{2}\tan\theta$
(C) $\tan\theta$
(D) $\frac{2}{3}\tan\theta$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. A uniform solid sphere of mass $M$ and radius $R$ is placed on a rough horizontal floor with coefficient of friction $\mu$. A constant horizontal force $F$ is applied directly at its center of mass. Choose the correct statement(s):
(A) The direction of static friction acting on the sphere at the contact point is opposite to the applied force $F$.
(B) The magnitude of static friction is $f_s = \frac{2}{7} F$.
(C) The linear acceleration of the center of mass during pure rolling is $a = \frac{5F}{7M}$.
(D) The maximum applied force $F$ for pure rolling without slipping is $F_{\text{max}} = \frac{7}{2}\mu M g$.
Q7. A uniform rod of mass $M$ and length $L$ is hinged smoothly at one end $O$ and held horizontally. It is released from rest. At the instant it is released: (Take upward/downward as vertical direction)
(A) The angular acceleration of the rod is $\alpha = \frac{3g}{2L}$.
(B) The linear acceleration of the center of mass of the rod is $a_{cm} = \frac{3}{4} g$.
(C) The vertical reaction force exerted by the hinge is $N_y = \frac{1}{4} M g$ directed upwards.
(D) The horizontal reaction force exerted by the hinge is zero.
Q8. A uniform circular disc of mass $M$ and radius $R$ is projected onto a rough horizontal floor with an initial forward linear velocity $v_0$ and a backward angular velocity $\omega_0$ (backspin). Let $\mu$ be the coefficient of kinetic friction. Which of the following statements is/are correct?
(A) If $\omega_0 R = 2v_0$, the disc will stop translating and rotating simultaneously, coming to a dead permanent stop.
(B) If $\omega_0 R > 2v_0$, the disc will reverse its linear direction and eventually roll backwards.
(C) If $\omega_0 R < 2v_0$, the disc will eventually execute pure rolling in the forward direction.
(D) Mechanical energy is conserved throughout the entire transition until pure rolling commences.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A uniform solid cylinder of mass $M = 5\text{ kg}$ and radius $R = 0.2\text{ m}$ is placed on a rough horizontal surface. A horizontal pulling force $F = 30\text{ N}$ is applied to the top edge of the cylinder. Assuming pure rolling without slipping, determine the acceleration of the center of mass of the cylinder in $\text{m/s}^2$.
Q10. A uniform rod of mass $M = 4.0\text{ kg}$ and length $L = 1.5\text{ m}$ is free to rotate in a horizontal plane about a fixed frictionless vertical pivot at one end $O$. A horizontal impulse of magnitude $J = 12.0\text{ N}\cdot\text{s}$ is delivered perpendicular to the rod at its Center of Percussion. Calculate the magnitude of the angular velocity $\omega$ of the rod immediately following the impact in $\text{rad/s}$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (B) | Torque about CM on frictionless table ($I_{cm} = \frac{1}{2}MR^2$) |
| Q2 | (C) | Work-Energy Theorem with Rolling ($K_{\text{total}} = \frac{7}{10}Mv^2$) |
| Q3 | (C) | Angular Impulse-Momentum about CM ($J d = I_{cm}\omega$) |
| Q4 | (A) | Angular Momentum of Projectile ($L = m v_x H_{\text{max}}$) |
| Q5 | (B) | Critical Incline Friction for Ring ($\mu \ge \frac{I_{cm}}{I_{cm}+MR^2}\tan\theta$) |
| Q6 | (A, B, C, D) | Force at CM of sphere: $f_s$ direction, acceleration, and threshold |
| Q7 | (A, B, C, D) | Hinged rod release: torque, CM acceleration, and vertical reaction |
| Q8 | (A, B, C) | Backspin sliding dynamics and IAOR angular momentum conservation |
| Q9 | 8 | Top-edge force on cylinder: $a_{cm} = \frac{4F}{3M} = 8\text{ m/s}^2$ |
| Q10 | 4 | Impact at Center of Percussion ($x = \frac{2}{3}L \implies \omega = 4\text{ rad/s}$) |
Detailed Explanations
Solution 1:
The table is smooth, so there is no friction force at the contact point. The only horizontal force acting on the disc is $F$ applied at the rim.
– Torque about the Center of Mass $CM$:
$$\tau_{cm} = F \cdot R$$
– The moment of inertia of a uniform circular disc about an axis perpendicular to its plane through the $CM$ is:
$$I_{cm} = \frac{1}{2} M R^2$$
– Applying the rotational equation of motion about the $CM$:
$$\tau_{cm} = I_{cm} \alpha \implies F R = \left(\frac{1}{2} M R^2\right) \alpha \implies \alpha = \frac{2F}{M R}$$
– Since the disc starts from rest and $F$ is constant:
$$\omega = \alpha t = \frac{2Ft}{MR}$$
Correct Answer: (B)
Solution 2:
Since the sphere rolls without slipping, static friction does zero work ($W_{f_s} = 0$). Hence, total mechanical energy is conserved.
– Total initial kinetic energy of the rolling solid sphere:
$$K_{\text{initial}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{2} I_{cm} \omega^2$$
For a solid sphere, $I_{cm} = \frac{2}{5} M R^2$, and with pure rolling $v = \omega R$:
$$K_{\text{initial}} = \frac{1}{2} M v^2 + \frac{1}{2}\left(\frac{2}{5} M R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2$$
– At maximum height $h$, the sphere comes momentarily to rest ($v = 0, \omega = 0$), so all kinetic energy is converted into gravitational potential energy:
$$M g h = \frac{7}{10} M v^2 \implies h = \frac{7v^2}{10g}$$
Correct Answer: (C)
Solution 3:
The rod is completely free on a smooth surface. During the instantaneous application of impulse $J$, the impulse exerts an angular impulse about the center of mass $CM$.
– Distance of line of action of impulse from $CM$: $d = L/4$.
– Angular impulse about $CM$:
$$\Delta L_{cm} = J \cdot d = J \left(\frac{L}{4}\right)$$
– The moment of inertia of the rod about its $CM$ is $I_{cm} = \frac{1}{12} M L^2$.
– Applying the angular impulse-momentum theorem:
$$I_{cm} \omega = J \left(\frac{L}{4}\right) \implies \left(\frac{1}{12} M L^2\right) \omega = \frac{J L}{4}$$
$$\omega = \frac{12}{4} \frac{J}{M L} = \frac{3J}{ML}$$
Correct Answer: (C)
Solution 4:
At the apex of the projectile trajectory:
– Vertical velocity $v_y = 0$.
– Horizontal velocity $v_x = u \cos\theta$.
– Maximum height above the launch point:
$$H_{\text{max}} = \frac{u^2 \sin^2\theta}{2g}$$
The velocity vector at the highest point is purely horizontal. The perpendicular distance from the launch origin to the horizontal line of velocity is precisely $H_{\text{max}}$.
Therefore, the magnitude of angular momentum about the origin is:
$$L = m v_x H_{\text{max}} = m (u \cos\theta) \left(\frac{u^2 \sin^2\theta}{2g}\right) = \frac{m u^3 \sin^2\theta \cos\theta}{2g}$$
Correct Answer: (A)
Solution 5:
For any rolling body down an incline of angle $\theta$, the required static friction to prevent slipping is:
$$f_s = \frac{M g \sin\theta}{1 + \frac{M R^2}{I_{cm}}}$$
For a thin ring, $I_{cm} = M R^2$:
$$f_s = \frac{M g \sin\theta}{1 + 1} = \frac{1}{2} M g \sin\theta$$
To prevent slipping, $f_s \le \mu_s N = \mu_s M g \cos\theta$:
$$\frac{1}{2} M g \sin\theta \le \mu_s M g \cos\theta \implies \mu_s \ge \frac{1}{2} \tan\theta$$
Thus, $\mu_{\text{min}} = \frac{1}{2}\tan\theta$.
Correct Answer: (B)
Solution 6:
Let $F$ be applied at the center of the sphere towards $+x$.
– If the floor were frictionless, the sphere would translate with acceleration $a = F/M$ without rotating ($\alpha = 0$). The bottom point would slide forward with acceleration $a$.
– Therefore, static friction $f_s$ must act backwards (towards $-x$) to produce a forward clockwise torque $\tau = f_s R$ to induce forward rotation. (Option A is correct).
– Equations of motion:
1. Translation: $F – f_s = M a$
2. Rotation: $f_s R = I_{cm} \alpha = \left(\frac{2}{5} M R^2\right) \alpha \implies f_s = \frac{2}{5} M (\alpha R)$
3. Pure rolling condition: $a = \alpha R$
Therefore, $f_s = \frac{2}{5} M a \implies M a = \frac{5}{2} f_s$.
Substitute into translation equation:
$$F – f_s = \frac{5}{2} f_s \implies F = \frac{7}{2} f_s \implies f_s = \frac{2}{7} F \quad \text{(Option B is correct).}$$
– Linear acceleration:
$$a = \frac{F – f_s}{M} = \frac{F – \frac{2}{7}F}{M} = \frac{5F}{7M} \quad \text{(Option C is correct).}$$
– Threshold for pure rolling without slipping ($f_s \le \mu M g$):
$$\frac{2}{7} F \le \mu M g \implies F \le \frac{7}{2} \mu M g \quad \text{(Option D is correct).}$$
All options (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
At the instant of release:
– Torque about fixed hinge $O$ is produced solely by gravity acting at $CM$ ($d = L/2$):
$$\tau_O = M g \left(\frac{L}{2}\right)$$
– Moment of inertia about hinge: $I_O = \frac{1}{3} M L^2$.
$$\tau_O = I_O \alpha \implies M g \frac{L}{2} = \left(\frac{1}{3} M L^2\right) \alpha \implies \alpha = \frac{3g}{2L} \quad \text{(Option A is correct).}$$
– Linear acceleration of the $CM$ (at distance $L/2$ from $O$):
$$a_{cm} = \alpha \left(\frac{L}{2}\right) = \left(\frac{3g}{2L}\right)\left(\frac{L}{2}\right) = \frac{3}{4} g \quad \text{(downward, Option B is correct).}$$
– Vertical force equation for the rod ($M g$ downward, vertical hinge reaction $N_y$ upward):
$$M g – N_y = M a_{cm} = M \left(\frac{3}{4} g\right) \implies N_y = M g – \frac{3}{4} M g = \frac{1}{4} M g \quad \text{(upward, Option C is correct).}$$
– Since the rod is released from rest, angular velocity $\omega = 0$. The centripetal acceleration of $CM$ along the horizontal direction is $a_{\text{centripetal}} = \omega^2 (L/2) = 0$. Hence, the horizontal reaction at the hinge $N_x = 0$. (Option D is correct).
All options (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
Let forward be $+x$ and clockwise be positive rotation. The disc has initial forward velocity $v_0 \hat{i}$ and counter-clockwise angular velocity $-\omega_0 \hat{k}$.
The contact point has velocity $v_{\text{contact}} = v_0 – (-\omega_0 R) = v_0 + \omega_0 R > 0$ forward, so kinetic friction acts backward: $\vec{f}_k = -\mu M g \hat{i}$.
– Linear acceleration: $a = -\mu g \implies v(t) = v_0 – \mu g t$.
– Torque about $CM$: $\tau = f_k R = \mu M g R$ (clockwise) $\implies \alpha = \frac{\tau}{I_{cm}} = \frac{\mu M g R}{\frac{1}{2} M R^2} = \frac{2\mu g}{R}$.
$$\omega(t) = -\omega_0 + \alpha t = -\omega_0 + \frac{2\mu g}{R} t$$
Pure rolling commences when $v(t) = \omega(t) R$:
$$v_0 – \mu g t = \left(-\omega_0 + \frac{2\mu g}{R} t\right) R = -\omega_0 R + 2\mu g t$$
$$v_0 + \omega_0 R = 3\mu g t \implies t_{\text{roll}} = \frac{v_0 + \omega_0 R}{3\mu g}$$
Velocity at the onset of pure rolling:
$$v_{\text{roll}} = v_0 – \mu g \left(\frac{v_0 + \omega_0 R}{3\mu g}\right) = v_0 – \frac{v_0 + \omega_0 R}{3} = \frac{2v_0 – \omega_0 R}{3}$$
– If $\omega_0 R = 2v_0 \implies v_{\text{roll}} = 0$ and $\omega_{\text{roll}} = 0$. The disc comes to a complete, permanent stop! (Option A is correct).
– If $\omega_0 R > 2v_0 \implies v_{\text{roll}} < 0$. The disc moves backward! (Option B is correct).
– If $\omega_0 R < 2v_0 \implies v_{\text{roll}} > 0$. The disc rolls forward! (Option C is correct).
– Kinetic friction performs negative work on the system, dissipating mechanical energy as heat; hence energy is NOT conserved. (Option D is incorrect).
Correct Answer: (A, B, C)
Solution 9:
Force $F$ is applied horizontally at the top rim of the cylinder ($r = +R$ above $CM$).
Let $a_{cm}$ be forward acceleration and $\alpha$ be clockwise angular acceleration.
Let static friction $f_s$ act forward ($+x$) at the bottom contact point.
– Translational equation:
$$F + f_s = M a_{cm} \quad \text{— (Equation 1)}$$
– Torque equation about $CM$:
$$\tau_{cm} = F R – f_s R = I_{cm} \alpha = \left(\frac{1}{2} M R^2\right) \alpha \implies F – f_s = \frac{1}{2} M (\alpha R) \quad \text{— (Equation 2)}$$
– Pure rolling condition: $a_{cm} = \alpha R$.
Therefore, Equation 2 becomes:
$$F – f_s = \frac{1}{2} M a_{cm} \quad \text{— (Equation 3)}$$
Add Equation 1 and Equation 3:
$$(F + f_s) + (F – f_s) = M a_{cm} + \frac{1}{2} M a_{cm}$$
$$2F = \frac{3}{2} M a_{cm} \implies a_{cm} = \frac{4F}{3M}$$
Substitute $F = 30\text{ N}$ and $M = 5\text{ kg}$:
$$a_{cm} = \frac{4(30)}{3(5)} = \frac{120}{15} = \mathbf{8\text{ m/s}^2}$$
Correct Answer: 8
Solution 10:
For a uniform rod of mass $M$ and length $L$ pivoted at one end $O$:
– The Center of Percussion is located at distance:
$$x_{cop} = \frac{I_O}{M d_{cm}} = \frac{\frac{1}{3} M L^2}{M (L/2)} = \frac{2}{3} L$$
With $L = 1.5\text{ m}$:
$$x_{cop} = \frac{2}{3}(1.5\text{ m}) = 1.00\text{ m}$$
The impulse $J = 12.0\text{ N}\cdot\text{s}$ is delivered at $x = 1.00\text{ m}$ perpendicular to the rod.
Applying angular impulse-momentum theorem about the fixed pivot $O$:
$$\Delta L_O = J \cdot x_{cop} = I_O \omega$$
where:
$$I_O = \frac{1}{3} M L^2 = \frac{1}{3}(4.0\text{ kg})(1.5\text{ m})^2 = \frac{1}{3}(4.0)(2.25) = 3.00\text{ kg}\cdot\text{m}^2$$
$$J \cdot x_{cop} = (12.0\text{ N}\cdot\text{s})(1.00\text{ m}) = 12.0\text{ kg}\cdot\text{m}^2/\text{s}$$
$$\omega = \frac{J x_{cop}}{I_O} = \frac{12.0}{3.00} = \mathbf{4\text{ rad/s}}$$
Correct Answer: 4