A hoop of mass M = 3 kg and radius R = 0.4 m rolls without slipping down a hill,

Question

A hoop of mass M = 3 kg and radius R = 0.4 m rolls without slipping down a hill, as shown in the figure. The lack of slipping means that when the center of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to v_CM, since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v - v = 0). Therefore, the angular speed of the rotating hoop is omega = V_CM/R. The initial speed of the hoop is v_i = 3 m/s, and the hill has a height h = 4.1 m. What is the speed Vf at the bottom of the hill? Replace the hoop with a bicycle wheel whose rim has mass M = 3 kg and radius R = 0.4 m, and whose hub has mass m = 2.1 kg, as shown in the figure. The spokes have negligible mass. What would the bicycle wheel's speed be at the bottom of the hill? (Assume that the wheel has the same initial speed and start at the same height as the hoop in part (a)).

Explanation / Answer

b) The kinetic energy of the rolling wheel is given by
  K = (1/2) Ip w2
Where Ip is the moment of inertia about an axis passing through the point p on the wheel which touches the ground
The moment of inertia of the about the centre is mR2 ,By parallel axis theorem the moment of inertia of a parallel axis at a distance R is given by
mR2 + mR2  = 2 mR2
But in this case of bicycle wheel there is one more mass attached at the hub.The hub has a mass of 2.1kg, the additional moment of inertia due to this mass is
I = MR2
The moment of inertia of the bycycle wheel at the point p will be the sum of these two
Ip  = 2mR2 + MR2   = (2m+M)R2
As in the previous case we have to equate the total energy at the two times
(1/2) Ip wi2 + mgh = (1/2) Ip wf2----------------------- 1
Where wi and wf are the initial and final angular velocities respectively and w = v / R
Substituting all this in equation 1
(1/2)(2m+M)R2 x (vi/R)2   + (m+M)gh = (1/2)(2m+M)R2 x (vf/R)2
m = 3kg, M = 2.1kg ,vi =3 m/s , h = 4.1 m , R = 0.4 m
(1/2)(2x3+2.1) x (3)2   + (3+2.1)x9.8x4.1 = (1/2)(2x3+2.1) x vf2
vf   = 7.72 m/s

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