A hoop of mass 1.2 kg and radius 75 cm has a string wrapped around its circumfer

Question

A hoop of mass 1.2 kg and radius 75 cm has a string wrapped around its circumference and lies flat on a horizontal frictionless table. The string is pulled with a constant force 3 N. How far does the center of the hoop travel in 1 s? Answer in units of m A hoop of mass 1.2 kg and radius 75 cm has a string wrapped around its circumference and lies flat on a horizontal frictionless table. The string is pulled with a constant force 3 N. How far does the center of the hoop travel in 1 s? Answer in units of m

Explanation / Answer

the string generates a torque on the hoop

the torque = Force x R where R is the radius of the hoop

torque = I alpha where I is the moment of inertia of the hoop and alpha is the angular accel

the moment of inertia of a hoop is MR^2 so we have

torque = FR= MR^2 alpha

or F=MR alpha

so alpha = F/MR = 3N/1.2kgx0.75m = 3N/0.9 kgm = 3.33rad/s/s

we get ang vel w, from w = alpha t = 3.33 rad/s/s x 1x =3.33 rad/s

the linear accelerations, a, is given by a=alpha R =3.33*0.75=2.4975 m/s/s

we recall from basic kinematics that an accelerating object will travel a distance:

dist = 1/2 at^2 = 1/2(2.4975m/s/s)(1s)^2=1.248m

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