A string is wrapped many times around a pulley and is connected to a block that is hanging vertically. The pulley consists of a wheel of radius 4 2 . 2 3 cm and mass 6 . 7 7 9 kg constructed with spokes that have negligible mass. The block accelerates downward

Physics Pulley and Motion Problems – Full Detailed Solution

Problem 1: Find the Mass of the Block Hanging from a Pulley

Given:

  • Radius of pulley, r = 42.23 cm = 0.4223 m
  • Mass of pulley, mp = 6.779 kg
  • Acceleration of block, a = 4.325 m/s²
  • Gravitational acceleration, g = 9.8 m/s²

To Find: Mass of the block, mb

Solution:

Torque, τ = Iα, where I = mp
Tension, T = mpa
Net force: mbg − T = mba

Substitute T into the equation:

mbg − mpa = mba

Rearranging the terms:

mb (g − a) = mp a

Solve for mb:

mb = (mp × a) / (g − a) = (6.779 × 4.325) / (9.8 − 4.325)
Mass of the block ≈ 5.36 kg

Problem 2: Distance to Place Cup to Catch a Rolling Ball on an Inclined Board

Given:

  • Length of board, L = 1.43 m
  • Width/Height of support, H = 12.3 cm = 0.123 m

To Find: Horizontal distance d from elevated end where the cup should be placed

Solution:

Determine the angle of inclination using tangent function:

tan θ = H / L = 0.123 / 1.43 ≈ 0.086 → θ ≈ 4.92°

Now, calculate horizontal displacement of the cup:

d = L × sin(θ) = 1.43 × sin(4.92°) ≈ 1.43 × 0.0858
Distance d ≈ 0.123 m

Problem 3: Time to Stop a Rotating Disk System

Given:

  • Disk A: m = 2.89 kg, r = 12.5 cm = 0.125 m
  • Disk B: m = 0.289 kg, r = 1.25 cm = 0.0125 m
  • Torque due to friction, τ = 0.233 Nm
  • Initial angular velocity, ω₀ = -6π rad/s

Solution:

Calculate the moment of inertia of both disks:

I = ½ m1 r1² + ½ m2 r2²
= ½ × 2.89 × (0.125)² + ½ × 0.289 × (0.0125)²
≈ 0.0226 + 0.00002 = 0.02262 kg·m²

Angular deceleration:

α = τ / I = 0.233 / 0.02262 ≈ 10.3 rad/s²

Using angular motion formula:

0 = ω₀ + αt ⇒ t = -ω₀ / α = 6π / 10.3 ≈ 1.83 s
Time to stop ≈ 1.83 s

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