Question

A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrapped around the axle, and the top is set spinning by applying T = 2.40 N of constant tension to each string. If it takes 0.740 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.

Answer 1

0.01051392 k·gm²/sec

Step-by-step explanation:

Torque, T = dL/dt

2Tr = L/t

L = 2Trt

L = 2(2.4)(0.00296)(0.74) = 0.01051392

Answer 2

Step-by-step explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let
`\alpha`be the angular acceleration

`2 T* r=I* \alpha`

`2* 2.4* 2.96* 10^(-3)=0.5* m* (2.96* 10^(-3))^2* \alpha`

`\alpha =(4* 2.4)/(m* 2.96* 10^(-3))`

`\alpha =(3.24* 10^3)/(m) rad/s^2`

`\omega =\omega _0+\alpha \cdot t`

`\omega =0+(3.24* 10^3)/(m)* 0.74`

`\omega =(2.4* 10^3)/(m) rad/s`

Angular momentum

`L=I\omega`

`L=0.5* mr^2* \omega`

`L=0.5* m* (2.96* 10^(-3))^2* (2.4* 10^3)/(m)`

`L=0.01051 kg-m^2/s`