you have a hoop and a uniform cylinder. they have the same mass m and radius r.
ID: 3897352 • Letter: Y
Question
you have a hoop and a uniform cylinder. they have the same mass m and radius r. You decide to race the objects from rest down an incline. both objects roll without slipping
1- which object will win the race ? explain
If they tie state that explicity?
2- next you roll the hoop back up the same incline. from the time you release the hope it rolls a distance d up the incline before turning around . draw an extended free body diagram for the hoop when it is rolling without slipping up the incline.
3- next , you make the incline very slippy (minimizing friction). You roll the hoop with the same initial velocity that it had ain part b. Will the hoop roll a distance greater than, less than, or equal to d before turning around ? explain
4- finally now that you have a very slippery incline you decide to race the hoop and cylinder from the rest down the incline again. which object will win the race ? explain. if they tie , state that explicity
Explanation / Answer
1
For same m and R, hoop will have more inertia.
For hoop I = mR^2
For cylinder I = 1/2*mR^2
KE = 1/2*Iw^2
KE_hoop = 1/2*mR^2*(v/R)^2 = 1/2*mv^2
KE_cyl = 1/2*(1/2*mR^2)*(v/R)^2 = 1/2*mv^2
Total KE_hoop = 1/2*mv^2 + 1/2*mv^2 = mv^2
Total KE_cyl = 1/2*mv^2 + 1/4*mv^2 = 3/4*mv^2
PE = mgh
For hoop:
mgh = mv^2
v = sqrt (gh)
For cyl:
mgh = 3/4*mv^2
v = sqrt (4gh/3)
V_cyl > v_hoop
Hence, the one with lower I will be faster. Cylinder will win.
2.
3.
As the friciton reduces, the tendency to slip will increase. In limiting case, the wheel will keep slipping at its own place.
Hence, distance d will reduce with reduced friction.
4.
Since now friciton is not there, there will be no rolling. Hence, only translational KE will be there (1/2 *mv^2)
For both, 1/2*mv^2 = mgh
v = sqrt (2gh)
Hence, there will be tie.
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