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 = 2.46 mm. Two strings are wrapped around the axle, and the top is set spinning by applying T = 3.40 N of constant tension to each string. If it takes 0.410 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.

(b) If the final tangential speed of point P, h = 29.0 mm above the ground, is 1.35 m/s and the angle is 17.0°, what is the top's moment of inertia?

Explanation / Answer

Given that

radius r=2.46 mm

tension T=3.4 n

time t=0.410 sec

basing on the relation b/w torque and angular momentum

now we find the angular momentum

torque=L/t

T*r=L/t

angular momentum L=T*t*r=3.4*2.46*10^-3*0.410=3.43*10^-3 kg*m^2/sec

now we find the finial tangential speed of point P

length L=29*10^-3/sin17=0.0992 m

finial speed v=[1.35^2+2*9.8*0.0992]^1/2

                    =1.941 m/s

now we find the moment of inertia

L=Ip*W

3.43*10^-3=Ip*(1.941/2.46*10^-3)

the top moment of inertia Ip=4.35*10^-6 kg.m^2

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