Release from rest a thin metal bicycle rim of diameter 0.1 cm and mass 9 kg from

Question

Release from rest a thin metal bicycle rim of diameter 0.1 cm and mass 9 kg from the down a 30° frictionless ramp that is 9 m long. (Note: acceleration down the ramp is g/4) (a) The translational velocity v when it reaches the bottom is v = ____ m/s. (b) The angular velocity when it reaches the bottom is = ____ rad/s. (c) The angular acceleration, = ______ rad/s2. (d) The percentage of the total KE that is rotational is ___ % (e) The angular momentum, at the bottom of the ramp is L = ______ kgm2/s

Explanation / Answer

r=0.05cm= 0.005m, m=9kg, g’=9/4 , d=9m, q= 300

a) Consider rim as a hoop,

Ihoop = MR2

Applying law of conservation of energy to the hoop along the ramp

KEi+PEi=KEf(kinetic + rotational) + PEf

1/2mvi2+mg’hi = KEf(kinetic + rotational)+mghf

1/2m(0)+mg’hi = KEf(kinetic + rotational) +mg(0)

mg’hi = KEf(kinetic + rotational)

mg’hi = 1/2mvf2 + 1/2Ihoopw2

m(g/4)hi = 1/2mvf2 + 1/2Ihoop(v/r)2

m(g/4)hi = 1/2mvf2 + 1/2mr2(vf/r)2

m(g/4)hi = 1/2mvf2 + 1/2mvf2

m(g/4)hi = mvf2

(g/4)hi = vf2

hi dsinq= 9sin30 = 4.5

vf = sqrt[ghi/4] = sqrt[(9.8*4.5)/4] = 3.32 m/s

b)

wf = vf/r = 3.31/0.005 = 662 rad/s

c) a=a*r     => a=a/r = (g/4)/r = (9.8/4)/0.005 = 490 rad/s2

d) KEtotal = KErot+KEtrans = ½(mr2) wf2 + 1/2mvf2 = ½*9*0.0052*6622 + ½*9*3.322 = 98.8 J

KErot = ½Iwf2 =½(mr2)wf2 = ½*9*0.0052*6622 = 49.3 J

KEtotal /KErot = (98.8 J/49.3J)*100 = 200%

e) Lf= Ihoop*wf = mr2*wf = 9*0.0052*662 = 0.15 kgm2s

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects