Register to reply 
Rotational Kinetic Energy 
Share this thread: 
#1
Mar608, 11:56 PM

P: 5

A bicycle has wheels of radius 0.33 m. Each wheel has a rotational inertia of 0.082 kg* m2 about its axle. The total mass of the bicycle including the wheels and the rider is 74 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
I thought this: Rotational KE = (1/2)Iw^2 =(1/2)(second bold number)w^2 Linear KE= (1/2)mv^2 = (1/2)(third bold number)(radius*w)^2 (i.e. plug in r*w for v) Total KE is equal to Rotational KE + Linear KE add the two eqns (1/2)Iw^2/ (some # * w^2) 


#2
Mar708, 12:06 AM

P: 68

i think you are correct
eventually one takes away the w^2 in both numerator and denominator, then gets a result independent of w 


#3
Mar708, 12:16 AM

P: 5

i keep getting .041/4.07 and that is not right



Register to reply 
Related Discussions  
Rotational Kinetic Energy  Introductory Physics Homework  2  
Rotational Kinetic Energy  Introductory Physics Homework  2  
Rotational kinetic energy  Introductory Physics Homework  1  
PLEASE HELP  Rotational Kinetic Energy  Introductory Physics Homework  6  
Rotational Kinetic Energy  Introductory Physics Homework  0 