Questions & explanations
1. A non-uniform ladder of length 5 m and mass 20 kg has its centre of mass 2 m from the bottom. It leans against a smooth vertical wall at 60° to the horizontal. The floor is rough with coefficient of friction 0.5. A person of mass 60 kg climbs up the ladder. What is the maximum distance (in m) the person can climb from the bottom before the ladder slips? (Take g = 10 m/s²)
- 3.0
- 4.0
- 3.5
- 4.5
Answer: 4.0
Using equilibrium conditions: horizontal forces: friction f = normal from wall N_w; vertical: normal from floor N_f = (M_L + m)g. Torque about bottom: N_w L sinθ = M_L g d_cm cosθ + m g d cosθ. At slip, f = μ N_f. Substituting and solving for d gives d = [μ (M_L + m) L tanθ - M_L d_cm] / m. With L=5, M_L=20, m=60, μ=0.5, θ=60°, tan60°=√3≈1.732, d_cm=2, we get d = 4.0 m.
2. A solid sphere rolls without slipping on a rough horizontal surface. A horizontal force F is applied at its centre. What is the direction of static friction?
- Upwards perpendicular to surface
- Opposite to F
- In the direction of F
- Zero
Answer: In the direction of F
For pure rolling, net torque about centre must be zero. Force F at centre produces zero torque. To prevent slipping, friction must provide a torque that opposes relative motion. If F is to the right, the bottom point tends to slip left; friction acts right (same as F) to oppose slip and provide clockwise torque, maintaining rolling.
3. A uniform ladder of length 5 m and weight 400 N leans against a smooth vertical wall at an angle of 60° with the horizontal. The floor is rough. What is the magnitude of the friction force at the floor required to keep the ladder in equilibrium?
- 200 N
- 115 N
- 230 N
- 400 N
Answer: 115 N
Taking torques about the foot, the weight (400 N) at 2.5 m from foot gives torque 400 × 2.5 × cos60° = 500 Nm clockwise. The wall normal force N_wall at height 5 sin60° = 4.33 m gives counterclockwise torque N_wall × 4.33. Equating gives N_wall = 115.5 N. Horizontal force balance gives friction f = N_wall = 115.5 N ≈ 115 N.
4. A solid cylinder of mass 2 kg and radius 0.1 m on a rough horizontal surface has a string wound around it. The string passes over a light frictionless pulley and is attached to a hanging block of mass 1 kg. The cylinder rolls without slipping. What is the acceleration of the hanging block? (g = 10 m/s²)
- 5.71 m/s²
- 4.00 m/s²
- 2.86 m/s²
- 3.33 m/s²
Answer: 4.00 m/s²
Let a be block acceleration, a_cm cylinder's. String constraint: a = 2 a_cm. For block: mg - T = ma. For cylinder: T - f = M a_cm, torque fR = Iα = (½MR²)(a_cm/R) => f = ½M a_cm. Substitute: T = M a_cm + ½M a_cm = (3/2)M a_cm. Then mg - (3/2)M a_cm = m(2 a_cm) => a_cm = mg/(3M/2 + 2m) = 10/(3+2)=2 m/s², so a = 4 m/s².
5. A block of base width 0.4 m and height 0.8 m, weight 100 N, is placed on a rough horizontal surface with μ = 0.3. A horizontal force F is applied at height y from the base. Which graph correctly shows the variation of the minimum force required to cause motion (sliding or toppling) as a function of y?
- A decreasing curve from 100 N at y=0 to 30 N at y=0.2 m, then constant at 30 N
- A horizontal line at 30 N for all y, then an increasing curve for y > 0.2 m
- A horizontal line at 30 N for all y, then a decreasing curve for y > 0.2 m
- A decreasing curve from 100 N at y=0 to 30 N at y=0.2 m, then a further decreasing curve below 30 N
Answer: A horizontal line at 30 N for all y, then a decreasing curve for y > 0.2 m
Sliding force F_s = μmg = 30 N constant. Toppling occurs when torque about edge: F * y = mg * (b/2) => F_t = (mg * b/2)/y = 20/y. For y ≤ 0.2 m, F_t ≥ 100 N > 30 N, so sliding governs (30 N). For y > 0.2 m, F_t < 30 N, so toppling governs (20/y decreasing). Hence graph: constant 30 N then decreasing curve.
6. A solid sphere slides with initial speed 7 m/s on a rough horizontal surface. What fraction of initial kinetic energy is lost before pure rolling begins?
- 5/7
- 2/7
- 1/2
- 1/4
Answer: 2/7
For a solid sphere, v_final = (5/7)v_0. Initial KE = ½Mv_0². Final total KE = ½Mv_final² + ½I_cm(v_final/R)² = ½Mv_final²(1 + I_cm/(MR²)) = ½Mv_final²(1 + 2/5) = (7/5)×½Mv_final². Substituting v_final² = (25/49)v_0² gives final KE = (7/5)×(25/49)×½Mv_0² = (5/7)×½Mv_0². Loss = 1 - 5/7 = 2/7 of initial KE.
7. A solid sphere rolls without slipping on a rough horizontal surface at speed v. It then enters a smooth horizontal surface. Which graph best shows its translational speed vs time as it moves from rough to smooth and then back to rough?
- Constant speed on rough, constant on smooth, constant on rough
- Constant speed on rough, increases on smooth, constant on rough
- Constant speed on rough, decreases on smooth, constant on rough
- Constant speed on rough, constant on smooth, decreases on rough
Answer: Constant speed on rough, constant on smooth, constant on rough
On rough surface, pure rolling with constant speed (no net force). On smooth surface, no friction, so no torque; sphere continues pure rolling with constant translational speed. On returning to rough, friction ensures pure rolling again at same constant speed. Thus speed remains constant throughout.
8. A solid sphere and a hollow sphere of same mass 1 kg and radius 0.1 m roll down an incline of length 2 m and angle 30°. Find the ratio of their translational kinetic energies at the bottom.
- 5 : 7
- 1 : 1
- 25 : 21
- 5 : 3
Answer: 25 : 21
For pure rolling, a = g sinθ/(1+I/(MR²)). Solid sphere: I/(MR²)=2/5, a_solid = (5/7)g sinθ = 25/7 m/s², v²_solid = 2aL = 100/7, translational KE = ½Mv² = 50/7 J. Hollow sphere: I/(MR²)=2/3, a_hollow = (3/5)g sinθ = 3 m/s², v²_hollow = 12, translational KE = 6 J. Ratio = (50/7):6 = 50:42 = 25:21.
9. A solid disc of mass 2 kg and radius 0.1 m rolls without slipping with angular speed 20 rad/s. What is its angular momentum about the contact point?
- 0.80 kg m²/s
- 0.20 kg m²/s
- 0.40 kg m²/s
- 0.60 kg m²/s
Answer: 0.60 kg m²/s
Angular momentum about contact point = I_cm ω + M R v_cm. For solid disc, I_cm = ½MR² = 0.01 kg m². v_cm = ωR = 2 m/s. So L = 0.01×20 + 2×0.1×2 = 0.2 + 0.4 = 0.6 kg m²/s. Alternatively, using parallel axis theorem: I_contact = I_cm + MR² = 0.03 kg m², L = I_contact ω = 0.03×20 = 0.6 kg m²/s.
10. In an Atwood machine, masses 4 kg and 2 kg hang from a disc pulley of mass 2 kg and radius 0.1 m. The system is released from rest. After the 4 kg mass falls 0.5 m, what is its speed? (g = 10 m/s²)
- √(10/3) m/s
- √(10/7) m/s
- √(20/7) m/s
- √(20/3) m/s
Answer: √(20/7) m/s
Energy conservation: loss in PE = (4-2)×10×0.5 = 10 J. Translational KE = ½×(4+2)v² = 3v². Rotational KE = ½×Iω², with I = ½×2×(0.1)² = 0.01 kg m², ω = v/0.1 = 10v, so rotational KE = 0.5v². Total KE = 3.5v² = (7/2)v². Equate: 10 = (7/2)v² → v² = 20/7 → v = √(20/7) m/s.
11. A solid sphere rolls without slipping down an incline of angle 30°. What is the minimum coefficient of static friction required? (g = 10 m/s²)
- tan30°/3.5
- tan30°/2.5
- tan30°/3
- tan30°/2
Answer: tan30°/3.5
For pure rolling, friction provides torque. Using Newton's second law and torque equation: mg sinθ - f = ma and fR = Iα with a = αR. For solid sphere, I = (2/5)mR². Solving gives f = (2/7)mg sinθ. Minimum μ = f/N = (2/7)mg sinθ / (mg cosθ) = (2/7) tanθ = tan30°/3.5.
12. A solid sphere is given an initial speed 7 m/s on a rough horizontal surface (μₖ = 0.5, g = 10 m/s²). How long does it take to start pure rolling?
- 0.4 s
- 0.2 s
- 0.5 s
- 0.7 s
Answer: 0.7 s
Using Newton's second law and rotational dynamics: friction provides linear deceleration a = μₖg and angular acceleration α = (5μₖg)/(2R). Pure rolling condition v = ωR gives t = v₀/(μₖg(1 + (I/mR²))) = v₀/(μₖg(1+2/5)) = (5v₀)/(7μₖg) = (5×7)/(7×0.5×10) = 0.7 s.