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

Question

A child's top is held in place, upright on a frictionless surface. The axle has a radius of r = 3.96 mm. Two strings are wrapped around the axle, and the top is set spinning by applying T = 2.65 N of constant tension to each string. If it takes 0.440 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.

^^^^^ I got this question right the second time (not 4.62) and the correct answer is 9.23X10-3 kg*m^2/s I just need help with the moment of inertia (the second question in the picture)

A child's top is held in place, upright on a frictionless surface. The axle has a radius of r 3.96 mm. Two strings are wrapped around the axle, and the top is set spinning by app constant tension to each string. If it takes 0.440 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 T=2.65 N of Number 4.62 kg. ms If the final tangential speed of point P h 25.0 mm above the ground, is 1.75 m/s and the angle is 20.0°, what is the top's moment of inertia? Number Top View: 0.105 kg m Incorrect.

Explanation / Answer

Solution: dL/dt = T.
For a constant torque T this integrates to

L(t) = L(0) + T t

The torque from the two strings is

T = 2 F R,

and the initial angular momentum L(0) is zero. Therefore

L(t) = 2 F R t

= 2 * 2.65N * 3.96E-3m * 0.440s
= 9.23E-3 kg 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) )

= 9.23E-3kgm^2/s * 2.50E-2m /( 1.75m/s*cos(20.0degrees) )
= 1.40E-4 kg m^2

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