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Question

A very light, inextensible string is wrapped around a cylindrical spool. The end of the string is held

fixed, and the spool is released so it starts falling, as the string unwinds. Because the spool is basically hollow,

you can take its moment of inertia to be I = mR2

(b) Find the linear acceleration of the spool as it accelerates toward the ground.

(c) Let the mass of the spool be 0.1 kg. What is its translational kinetic energy after it has fallen for 0.5 s?

(d) What is its rotational kinetic energy at that time?

(e) What is the tension in the string? Does this change as the spool falls? (Remember the mass of the

string is negligible.)

(f) If the radius of the spool is 3 cm, what is the magnitude of the torque (around the center of mass) exerted

by the tension?

Answer 1

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b) The torque created by the weight of the spool about the last point of contact with the string is

τ = mgR

But also

τ = Iα

BUT about the point of contact, (by the parallel axis theorem),

I = mR² + mR² = 2mR²

so mgR = 2mR²α

means that α = g/(2R)

and so a = αR = g/2 ≈ 4.9 m/s²

c) After 0.5 s, v = a*t = 4.9m/s² * 0.5s = 2.45 m/s

translational KE = ½ * 0.1kg * (2.45m/s)² = 0.3 J

d) After 0.5 s, the spool has fallen

h = ½*4.9m/s²*(0.5s)² = 0.61 m

and so has lost PE = 0.1kg * 9.8m/s² * 0.61m = 0.6 J

and so the rotational KE must be 0.3 J

e) T = m(g - a) = 0.1kg * 4.9m/s² = 0.49 N

f) τ = 0.49N * 0.03m = 0.015 N·m

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