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Block lowering a drum, energy method

Short answer

Work-energy: 10×9.81×2 = ½ v²(10 + 1.25/0.3²) → v = 4.05 m/s. Acceleration a = v²/(2h) = 4.11 m/s²; angular acceleration α = 13.69 rad/s²; rope tension T = 10(9.81 - 4.11) = 57.0 N.

A drum of mass 20 kg, radius 0.30 m and radius of gyration 0.25 m about its fixed horizontal axis has a light rope wound round it carrying a 10 kg block at the free end. The system is released from rest. Using the work-energy method, find the speed of the block after it has descended 2 m. Hence find the acceleration of the block, the angular acceleration of the drum and the tension in the rope.

Given: drum M=20M = 20 kg, r=0.30r = 0.30 m, k=0.25k = 0.25 m; block m=10m = 10 kg; fall h=2h = 2 m from rest.

      ___
     /   \  drum (fixed axle)
     \___/|
          | rope
         [m]  falls h = 2 m

Drum moment of inertia

I=Mk2=20(0.25)2=1.25 kg⋅m2Ir2=13.889 kgI = Mk^2 = 20(0.25)^2 = 1.25\ \text{kg·m}^2 \qquad \frac{I}{r^2} = 13.889\ \text{kg}

Work-energy method

Rope tension does no net work on the system (it acts on both the block and the drum with opposite signs); only the weight of the block does work:

U=mgh=10(9.81)(2)=196.2 JU = mgh = 10(9.81)(2) = 196.2\ \text{J}

Final kinetic energy, with ω=v/r\omega = v/r:

T2=12mv2+12Iω2=12v2(m+Ir2)=12v2(10+13.889)T_2 = \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 v^2\left(m + \frac{I}{r^2}\right) = \tfrac12 v^2(10 + 13.889) 0+196.2=12v2(23.889)  ⇒  v=4.053 m/s0 + 196.2 = \tfrac12 v^2(23.889) \;\Rightarrow\; v = 4.053\ \text{m/s}

Acceleration of the block

For constant acceleration from rest, v2=2ahv^2 = 2ah:

a=v22h=(4.053)24=4.107 m/s2a = \frac{v^2}{2h} = \frac{(4.053)^2}{4} = 4.107\ \text{m/s}^2

Check: a=mgm+I/r2=98.123.889=4.107a = \dfrac{mg}{m + I/r^2} = \dfrac{98.1}{23.889} = 4.107 m/s².

Angular acceleration and tension

α=ar=4.1070.3=13.69 rad/s2T=m(g−a)=10(9.81−4.107)=57.03 N\alpha = \frac{a}{r} = \frac{4.107}{0.3} = 13.69\ \text{rad/s}^2 \qquad T = m(g - a) = 10(9.81 - 4.107) = 57.03\ \text{N}

Answer: v = 4.05 m/s after 2 m; a = 4.11 m/s²; α = 13.69 rad/s²; T = 57.0 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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