I dont know how to work it through.
time period T = 2*pi* (physical pendulum formula)
I = moment of inertia of ring about point of suspension = mr2 + mr2 = 2*m*r2
length l = distance of point of suspension from center of gravity i.e from center = r
so T = 2*pi* = 2*pi*
angular frequency = 2*pi/T = = = 0.97 rad/s
I dont know how to work it through. 26. A hoop (i.e., a ring) of radius...
Exercise 10.20A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1 ) ). After the hoop has descended 65.0 cm , calculate Part A the angular speed of the rotating hoop and w= rad/sPart B the speed of its center. v= m/s
A string is attached to the rim of a small hoop of radius r= 8.00×10−2 m and mass m = 0.180 kg and then wrapped several times around the rim. If the free end of the string is held in place and the hoop is released from rest and allowed to drop, as shown in the figure (Figure 1) , calculate the angular speed and the translational speed of the rotating hoop after it has descended h = 0.750 m...
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