Solution Week 87 (5/10/04)
Leaving the hemisphere Assume that the particle slides o to the right. Let vx and vy be its horizontal and vertical velocities, with rightward and downward taken to be positive, respectively. Let Vx be the velocity of the hemisphere, with leftward taken to be positive. Conservation of momentum gives mvx = M Vx = Vx = m M vx . (1)
Consider the moment when the particle is located at an angle down from the top of the hemisphere. Locally, it is essentially on a plane inclined at angle , so the three velocity components are related by vy = tan vx + V x = vy = tan 1 + m M vx . (2)
To see why this is true, look at things in the frame of the hemisphere. In this frame, the particle moves to the right with speed vx + Vx , and downward with speed vy . Eq. (2) represents the constraint that the particle remains on the hemisphere, which is inclined at an angle at the given location. Let us now apply conservation of energy. In terms of , the particle has fallen a distance R(1 cos ), so conservation of energy gives 1 1 2 2 2 m(vx + vy ) + M Vx = mgR(1 cos ). 2 2
2 to obtain Using eqs. (1) and (2), we can solve for vx 2 vx =
(3)
2gR(1 cos ) (1 + r) 1 + (1 + r) tan2
where r
m . M
(4)
This function of starts at zero for = 0 and increases as increases. It then achieves a maximum value before heading back down to zero at = /2. However, vx cannot actually decrease, because there is no force available to pull the particle to the left. So what happens is that vx initially increases due to the non-zero normal force that exists while contact remains. But then vx reaches its maximum, which corresponds to the normal force going to zero and the particle losing contact with the hemisphere. The particle then sails through the air with constant vx . Our goal, 2 in eq. (4) is maximum. Setting the then, is to nd the angle for which the vx derivative equal to zero gives 0 = = = = 0 = 2 tan cos2 1 + (1 + r) tan2 cos3 2(1 + r)(1 cos ) 1 + (1 + r) tan2 sin (1 cos )(1 + r) (5)
0 = cos3 + (1 + r)(cos cos3 ) 2(1 + r)(1 cos ) 0 = r cos3 3(1 + r) cos + 2(1 + r). 1
This is the desired equation that determines . It is a cubic equation, so in general it cant be solved so easily for . But in the special case of r = 1, we have 0 = cos3 6 cos + 4. By inspection, cos = 2 is an (unphysical) solution, so we nd (cos 2)(cos2 + 2 cos 2) = 0. The physical root of the quadratic equation is cos = 3 1 0.732 = 42.9 . (7) (6)
(8)
Alternate solution: In the reference frame of the hemisphere, the horizontal speed of the particle vx + Vy = (1 + r)vx . The total speed in this frame equals this horizontal speed divided by cos , so (1 + r)vx . (9) cos The particle leaves the hemisphere when the normal force goes to zero. The radial F = ma equation therefore gives v= mv 2 . (10) R You might be concerned that we have neglected the sideways ctitious force in the accelerating frame of the hemisphere. However, the hemisphere is not accelerating beginning at the moment when the particle loses contact, because the normal force has gone to zero. Therefore, eq. (10) looks exactly like it does for the familiar problem involving a xed hemisphere; the dierence in the two problems is in the calculation of v . Using eqs. (4) and (9) in eq. (10) gives mg cos = mg cos = Simplifying this yields 1 + (1 + r) tan2 cos3 = 2(1 + r)(1 cos ), (12) 2gR(1 cos ) m(1 + r)2 . 2 R cos (1 + r) 1 + (1 + r) tan2 (11)
which is the same as the second line in eq. (5). The solution proceeds as above.
Remark: Lets look at a few special cases of the r m/M value. In the limit r 0 (in other words, the hemisphere is essentially bolted down), eq. (5) gives cos = 2/3 0 = cos3 3 cos + 2 = 48.2 , 0 = (cos 1)2 (cos + 2). (13)
a result which may look familiar to you. In the limit r , eq. (5) reduces to = (14)
Therefore, = 0. In other words, the hemisphere immediately gets squeezed out very fast to the left. For other values of r, we can solve eq. (5) either by using the formula for the roots of a cubic equation (very messy), or by simply doing things numerically. A few numerical results are:
r 0 1/2 1 2 10 100 1000
cos .667 .706 .732 .767 .858 .947 .982 1
48.2 45.1 42.9 39.9 30.9 18.8 10.8 0
