b) When the weight has fallen a distance 75.3 cm, what is the kinetic energy of

Question

b) When the weight has fallen a distance 75.3 cm, what is the kinetic energy of the rotating wheel?

A wheel rotates about a horizontal axis, as shown in the figure. It has an outer radius R1 = 18.8 cm and mass M1 = 4.51 kg. Attached to it concentrically is a hub (shown cross-hatched) of radius R2 = 5.26 cm and mass M2 = 1.16 kg. The axle has negligible radius and mass. Both the wheel and the hub are solid and have uniform density. Suspended from a massless string that is wound around the hub is a weight of mass m = 512 g.

a) What is the acceleration of the hanging weight after it is released?
  m/s2

b) When the weight has fallen a distance 75.3 cm, what is the kinetic energy of the rotating wheel?

  J

Explanation / Answer

mg-T=ma

and T*0.0526=(I1+I2)*a/0.0526

T*0.0526^2=(4.51*0.188^2+1.16*0.0526^2)*a

or T=58.77a

substituting in the first equation we get

0.512*9.8-58.77a=0.512a

a=0.16782m/s2

2)final velocity=(2*0.16782*0.753)^0.5=0.5027m/s

w=a/R=0.5027/0.0526=9.557rad/s

KE=1/2*I*w^2=1/2*4.51*0.188^2/2*9.557^2

KE=3.639J

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