A child\\\'s top is held in place, upright on a frictionless surface. The axle h

Question

A child's top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.21 mm. Two strings are wrapped around the axle, and the top is set spinning by applying T = 2.90 N of constant tension to each string. If it takes 0.260 s for the string to unwind, how much angular momentum does the top acquire? Assume that the strings do not slip as the tension is applied.

If the final tangential speed of point P, h = 33.0 mm above the ground, is 1.65 m/s and the angle is 18.0°, what is the top's moment of inertia?

Explanation / Answer

Ans:-

Well, the torque applied is
= F • r = 2 * 2.9N * 0.00221m = 0.013 N·m
But = dL/dt, so
dL = dt = 0.013N·m * 0.260s = 3.33*10^-3 kg·m²/s
dL is the change in angular momentum; since Li = 0, Lf = dL

Lf = 3.33*10^-3kg.m^2/s

For the second part:
h = r cos(theta), so r = h/cos(theta)
Furthermore,
v = w r, so
w = v/r = v / (h/cos(theta)) = v cos(theta)/h
Therefore
L = I w gives
I = L/w = L h /( v cos(theta) )

= 3.33*10^-3kgm^2/s * 3.30*10^-2m /( 1.65m/s*cos(18.0degrees) )
= 7*10^-5 kg m^2

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