Advanced Rotational Dynamics Problem

Problem Statement 🔴

A uniform solid cylinder of mass $M$ and radius $R$ is placed on a rough horizontal surface with coefficient of friction $\mu = 0.3$. A light, inextensible string is wound around the cylinder and passes over a frictionless pulley. A block of mass $m = \frac{M}{2}$ is attached to the free end of the string.

When the system is released from rest:

1. Find the acceleration of the block
2. Find the angular acceleration of the cylinder
3. Find the time when pure rolling begins
4. Find the velocity of the block when pure rolling begins

Given: $M = 4$ kg, $R = 0.5$ m, $g = 10$ m/s²

Solution 1: Force and Torque Analysis

Step 1: Free Body Diagram Analysis

For the block (mass m):

• Tension $T$ upward
• Weight $mg$ downward
• Acceleration $a$ downward

Equation: $mg - T = ma$ → $T = mg - ma$ … (1)

For the cylinder (mass M):

• Tension $T$ at the point of contact (rightward)
• Friction $f$ at the bottom (leftward initially)
• Normal reaction $N$ upward
• Weight $Mg$ downward
• Linear acceleration $a_c$ rightward
• Angular acceleration $\alpha$ clockwise

Equations:

• $T - f = Ma_c$ … (2)
• $TR - fR = I\alpha$ … (3) where $I = \frac{1}{2}MR^2$ for solid cylinder
• $N = Mg$ … (4)

Step 2: Constraint Relations

Before pure rolling begins:

• The string unwinds: $v = a_ct + R\omega t$
• Acceleration relation: $a = a_c + R\alpha$ … (5)

Step 3: Solving the Equations

From (3): $TR - fR = \frac{1}{2}MR^2\alpha$ $T - f = \frac{1}{2}MR\alpha$ … (6)

From (2) and (6): $Ma_c = \frac{1}{2}MR\alpha$ $a_c = \frac{1}{2}R\alpha$ … (7)

From (5): $a = \frac{1}{2}R\alpha + R\alpha = \frac{3}{2}R\alpha$ $\alpha = \frac{2a}{3R}$ … (8)

From (7): $a_c = \frac{1}{2}R \times \frac{2a}{3R} = \frac{a}{3}$ … (9)

Substituting in (1): $T = mg - m\left(\frac{3a_c}{\text{from constraint}}\right)$

Using the string constraint: $a = 3a_c$

From (1): $T = mg - ma = mg - 3ma_c$ … (10)

From (2): $T - f = Ma_c$ $f = T - Ma_c = mg - 3ma_c - Ma_c = mg - (3m + M)a_c$ … (11)

For no slipping condition: $f \leq \mu N = \mu Mg$ $mg - (3m + M)a_c \leq \mu Mg$

Given $m = \frac{M}{2}$: $Mg - (3 \times \frac{M}{2} + M)a_c \leq \mu Mg$ $Mg - \frac{5M}{2}a_c \leq \mu Mg$ $1 - \frac{5}{2}\frac{a_c}{g} \leq \mu$

Substituting values: $1 - \frac{5}{2}\frac{a_c}{10} \leq 0.3$ $1 - \frac{a_c}{4} \leq 0.3$ $\frac{a_c}{4} \geq 0.7$ $a_c \geq 2.8$ m/s²

This means the friction force is insufficient to prevent slipping initially.

Step 4: Initial Motion (With Slipping)

During slipping: $f = \mu N = \mu Mg = 0.3 \times 4 \times 10 = 12$ N

From (2): $T - 12 = 4a_c$ From (10): $T = 20 - 2a$ (since $m = 2$ kg)

Using constraint $a = 3a_c$: $T = 20 - 2(3a_c) = 20 - 6a_c$

Substituting in $T - 12 = 4a_c$: $20 - 6a_c - 12 = 4a_c$ $8 = 10a_c$ $a_c = 0.8$ m/s²

Therefore:

• $a = 3a_c = 2.4$ m/s²
• $\alpha = \frac{2a}{3R} = \frac{2 \times 2.4}{3 \times 0.5} = 3.2$ rad/s²

Step 5: Time When Pure Rolling Begins

Condition for pure rolling: $v = R\omega$

Using kinematics:

• $v = at = 2.4t$
• $\omega = \alpha t = 3.2t$

Setting $v = R\omega$: $2.4t = 0.5 \times 3.2t$

This condition is not satisfied, indicating our assumption needs refinement.

Let’s use energy approach or reconsider the constraint.

Actually, the correct constraint during rolling: The relative velocity at contact point = 0 $v_{cylinder} - R\omega = v_{block} - \text{string extension rate}$

This is complex. Let’s try Solution 2.

Solution 2: Energy and Conservation Approach

Step 1: Initial Configuration Analysis

Total energy is conserved (no non-conservative forces doing work).

Initial energy: $E_i = 0$ (taking reference at initial position)

At any time $t$:

• Block descends by distance $s = \frac{1}{2}at^2$
• Cylinder translates by $x = \frac{1}{2}a_ct^2$
• Cylinder rotates by angle $\theta = \frac{1}{2}\alpha t^2$

Energy equation: $mg\frac{1}{2}at^2 = \frac{1}{2}mv^2 + \frac{1}{2}Mv_c^2 + \frac{1}{2}I\omega^2$

Substituting expressions: $mg\frac{1}{2}at^2 = \frac{1}{2}m(at)^2 + \frac{1}{2}M(a_ct)^2 + \frac{1}{2} \times \frac{1}{2}MR^2(\alpha t)^2$

Simplifying: $mga = ma^2 + Ma_c^2 + \frac{1}{2}MR^2\alpha^2$

Using constraints $a = 3a_c$ and $\alpha = \frac{2a}{3R}$: $Mg \times 3a_c = m(3a_c)^2 + Ma_c^2 + \frac{1}{2}MR^2\left(\frac{6a_c}{3R}\right)^2$ $3Mga_c = 9ma_c^2 + Ma_c^2 + 2Ma_c^2$ $3Mga_c = 9ma_c^2 + 3Ma_c^2$

Given $m = \frac{M}{2}$: $3Mga_c = 9 \times \frac{M}{2}a_c^2 + 3Ma_c^2$ $3g = \frac{9}{2}a_c + 3a_c$ $3g = \frac{15}{2}a_c$ $a_c = \frac{6g}{15} = \frac{2g}{5} = 4$ m/s²

Therefore:

• $a = 3a_c = 12$ m/s²
• $\alpha = \frac{2a}{3R} = \frac{24}{1.5} = 16$ rad/s²

Step 2: Verification with Friction Constraint

Maximum available friction: $f_{max} = \mu Mg = 0.3 \times 40 = 12$ N

Required friction for these accelerations: From earlier equation: $f = mg - (3m + M)a_c = 20 - (3 + 4) \times 4 = 20 - 28 = -8$ N

The negative sign indicates friction acts opposite to our assumed direction. This is acceptable as it’s within the limit.

Step 3: Time When Pure Rolling Begins

Condition: $v_c = R\omega$ $a_ct = R\alpha t$ $a_c = R\alpha = 0.5 \times 16 = 8$ m/s²

But we found $a_c = 4$ m/s². This inconsistency suggests the problem needs careful analysis.

Let me reconsider the problem setup.

Solution 3: Correct Approach - Two-Phase Motion

Phase 1: Initial Slipping (0 ≤ t ≤ t₁)

During this phase, kinetic friction acts.

Forces on cylinder:

• $T - \mu Mg = Ma_c$ … (1)
• $TR + \mu MgR = I\alpha$ … (2)

For block: $mg - T = ma$ … (3)

Constraints:

• String constraint: $a = a_c + R\alpha$ … (4)

Substituting $I = \frac{1}{2}MR^2$ in (2): $TR + \mu MgR = \frac{1}{2}MR^2\alpha$ $T + \mu Mg = \frac{1}{2}MR\alpha$ … (5)

From (1): $T = Ma_c + \mu Mg$ … (6)

Substituting (6) in (5): $Ma_c + \mu Mg + \mu Mg = \frac{1}{2}MR\alpha$ $Ma_c + 2\mu Mg = \frac{1}{2}MR\alpha$ … (7)

From (3) using (6): $mg - (Ma_c + \mu Mg) = ma$ $mg - Ma_c - \mu Mg = ma$ … (8)

Using constraint (4): $a = a_c + R\alpha$

Substituting in (8): $mg - Ma_c - \mu Mg = m(a_c + R\alpha)$ $mg - Ma_c - \mu Mg = ma_c + mR\alpha$

Grouping terms: $mg - \mu Mg = (M + m)a_c + mR\alpha$ … (9)

From (7): $R\alpha = \frac{2a_c}{M}(Ma_c + 2\mu Mg)$

Substituting in (9): $mg - \mu Mg = (M + m)a_c + m \times \frac{2a_c}{M}(Ma_c + 2\mu Mg)$

This is quadratic in $a_c$. Let’s substitute numerical values:

$20 - 12 = (4 + 2)a_c + 2 \times \frac{2a_c}{4}(4a_c + 24)$ $8 = 6a_c + a_c(4a_c + 24)$ $8 = 6a_c + 4a_c^2 + 24a_c$ $4a_c^2 + 30a_c - 8 = 0$

Solving quadratic equation: $a_c = \frac{-30 \pm \sqrt{900 + 128}}{8} = \frac{-30 \pm \sqrt{1028}}{8}$

Taking positive root: $a_c = \frac{-30 + 32.06}{8} = 0.258$ m/s²

Therefore:

• $a_c = 0.258$ m/s²
• $R\alpha = \frac{2 \times 0.258}{4}(4 \times 0.258 + 24) = 3.2$ m/s²
• $\alpha = \frac{3.2}{0.5} = 6.4$ rad/s²
• $a = a_c + R\alpha = 0.258 + 3.2 = 3.458$ m/s²

Phase 2: Pure Rolling (t ≥ t₁)

When $v_c = R\omega$, pure rolling begins.

Time when this occurs: $v_c = a_ct = R\omega = R\alpha t$ $a_c = R\alpha$

This gives $0.258 = 3.2$, which is impossible.

Conclusion: The cylinder will never achieve pure rolling in this configuration. The friction is insufficient to ever satisfy the rolling condition.

Final Answer

After careful analysis, the correct results are:

1. Block acceleration: $a = 3.46$ m/s² (downward)
2. Cylinder acceleration: $a_c = 0.258$ m/s² (rightward)
3. Angular acceleration: $\alpha = 6.4$ rad/s² (clockwise)
4. Pure rolling: Never achieved (friction insufficient)

Key Concepts Used

• Newton’s laws of motion
• Rotational dynamics
• Friction and rolling conditions
• Constraint relations
• Energy conservation principles

Common Mistakes

1. Incorrect constraint relations: Not properly relating linear and angular accelerations
2. Friction direction: Assuming wrong direction for friction force
3. Rolling condition: Misapplying pure rolling condition
4. Sign conventions: Inconsistent sign conventions for forces and accelerations

Related Problems

1. Same setup with $\mu = 0.6$ (analyze when pure rolling occurs)
2. Replace solid cylinder with hollow cylinder
3. Add an additional mass to the system
4. Consider the case where the string is wound at radius R/2

Video Solution

[Link to detailed video explanation covering all three solution approaches]

Time Estimate: 12-15 minutes for complete solution Difficulty Level: Very Challenging 🔴 Success Rate: ~20% in first attempt Key Insight: Recognize when pure rolling is not achievable