A thin, light wire is wrapped around the rim of a wheel, as shown in the followi
ID: 1630190 • Letter: A
Question
A thin, light wire is wrapped around the rim of a wheel, as shown in the following figure. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius 0.252 m. An object of mass 4.30 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.(Figure 1.) If the suspended object moves downward a distance of 3.40 m in 1.96 s, what is the mass of the wheel? Express your answer with the appropriate units.Explanation / Answer
Given
wheel uniform disc of radius r = 0.252 m
mass of the object 4.30 kg
distance moved downward is 3.4 m in time t = 1.96 s
here linear kinetic energy of object turns to rotate the wheel which is rotational kinetic energy
the wheel starts from rest and the object also
energy of the object from Potential energy to Kinetic energy is Mgh = 4.3*9.8*3.4 J = 143.276J
now we can calculate the acceleration of the object at it is at 3.4 m and t=1.96 s is
using equations of motion s= ut +0.5*at^2 ==> a = (s-ut)/(0.5*t^2) m/s2
a = (3.4-0*1.96)/(0.5*1.96^2) m/s2 = 1.77 m/s2
the velocityo f the object at position 3.4 m is (final velocity ) using equations of motion V = u+at = 0+1.77*1.96 m/s = 3.4692 m/s
the translatory kinetic energy of object is k.e1 = 0.5*M*v^2 = 0.5*4.3*3.4692^2J = 25.88 J
the rotational kinetic energy of the wheel is = 143.276 - 25.88 J = 117.396 J
The rotational kineitic enerty is K.E_r = 0.5*I*W^2
w is angular speed W = v/r = 3.4692/0.252 rad.s = 13.77 rad/s
and the moment of inertia of uniform disc is I = 0.5*M*r^2
the rotationsl kinetic energy is = 0.5*I*W^2
117.396 = 0.5*(0.5*m*0.252^2)(13.77^2)
mass of the object is M = 38.998 kg = 39 kg
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