Suppose a round rigid body (hoop, sphere, disk) with mass m, radius R, and radiu

Question

Suppose a round rigid body (hoop, sphere, disk) with mass m, radius R, and radius of gyration kG rolls without sliding toward a small obstruction ("bump"). The obstruction has a height h above the flat, even surface. In this problem, you receive step-by-step instructions how to use conservation of angular momentum to find the minimum velocity VG that the body must have in order to roll over the obstruction at A The linear impulse-momentum diagram is given above at the right There is no impulse due to a normal force because the body has lifted off the surface. Why has the impulse due to the weight been ignored? Find the angular momentum about point A just before (I G omega plus the moment of mvG) and immediately after (IG omega * + the moment of mvG*). This quantity is conserved, so equate the two expressions. Note that immediately after the collision, A is the ICOR for the body. Use that to relate vG * to omega * and use "rolling without sliding" to relate vG and omega. Finally, although energy is not conserved from start to finish, it is conserved from immediately after the impact on. To just clear the obstruction, the kinetic energy immediately after impact must equal the potential energy as the body clears the obstruction (in other words, the velocity is zero as the body reaches its highest point). Use that to solve for the minimum value of vG. Use your expression from (d) to find the minimum velocity for a disk of mass m = 2.5 kg, radius R = 0.3 m and an obstruction height h = 0.05 m.

Explanation / Answer

a) Impluse due to weight is ignored because it has no involvement in the horizontal motion of the object. Also it is very negligible and can be ignored generally.

b) angular momentum of object about point A Immediately before = Ig.w + m.Vg.R ; Ig= moment of interia of object about its CM.

  angular momentum of object about point A Immediately after = Ig.w* + m.Vg*.R

by principle of conservation of angular momentum ; Ig.w + m.Vg.R  = Ig.w* + m.Vg*.R

c) As A is the ICOR; we get a relation Vg* = w* .R

also from rolling without slipping we get ; Vg = wR;

d) the energy equation will be ; m.g.h = 0.5 . Ig . (w*)^2 + 0.5 . m .(Vg*)^2 , from principle of conservation of energy.

solving this we get, m.g.h = 0.5 . Ig . (Vg*/R)^2 + 0.5 . m .(Vg*)^2

(Vg*)^2 =  2.m.g.h/( Ig /(R)^2 + m) Ig = m.(Kg )^2 ; where Kg = radius of gyration.

(Vg*)^2 =  2.g.h/((Kg/R)^2 + 1) ;

Vg* = sqrt[2.g.h/((Kg/R)^2 + 1)] ;

e) for disk ; Kg= R/sqrt(2) ;

putting the values given , we get Vg* = 0.8083 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.