Solutions Ph1a 6
Chefung Chan ([e-mail]), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead.
6.1
6.1.a
QP7 (5 points)
(1 point)
The total kinetic energy just before the collision is equal to the total potential energy before the masses are released. K = Um + UM = (m + M )gh 6.1.b (2 points) 2gh and
For inelastic collisions, momentum is conserved and kinetic energy is not conserved. Define v0 = the velocity after the collision as vf . Conservation of momentum gives mv0 + M (-v0 ) = (m + M )vf vf = The resulting kinetic energy (Kf ) is Kf = 1 (m - M )2 2 (m + M )vf = gh. 2 (m + M ) m-M v0 m+M
Defining the maximum height after the collision as h , conservation of energy (after the collision) gives (m + M )gh = (m - M )2 gh (m + M )
2
h = 6.1.c (2 points)
m-M m+M
h
Elastic collisions conserve both momentum and kinetic energy. Defining the velocities of the masses after the collision as vm and vM gives final momentum and kinetic energy of pf = mvm + M vM Kf = The conservation laws give 1 1 1 2 2 2 (m + M )v0 = mvm + M vM 2 2 2 Solving these equations for the velocity of the small mass vm = Km = 1 mv 2 = 2 m m - 3M v0 m+M m - 3M m+M 1
2
1 1 2 2 mvm + M vM 2 2
(m - M )v0 = mvm + M vM
(mgh)
Conserving energy for the small mass after the collision gives the maximum height h mgh = m - 3M m+M m - 3M m+M
2
(mgh)
2
h =
h
6.2
QP12 (5 points)
Z
R
r
0
6.2.a
(3 points)
The force diagram for the marble is...
N Fc
Fg
Balancing the forces at equilibrium gives: X : N sin = Fc = m 2 r0 Y : N cos = Fg = mg By dividing these two equations, we get the angle of the normal force: tan = 2 r0 g 2 r0 ] g
= tan-1 [
2
By looking at the triangle in the original diagram, we see: sin = N= r0 R
m 2 r0 sin
N = m 2 R 6.2.b (1 point)
Again by looking at the triangle, we can get another expression for tan , in addition to the one from part a. tan = r0
2 R2 - r0
=
2 r0 g
1 4 = 2 2 R2 - r0 g r0 = 6.2.c (1 point) R2 - g2 4
g When < R , the above answer for r0 is imaginary. If placed at the bottom, it will remain there. However, if given any initial motion, the particle will not settle into a circular orbit but "slosh" about in the bowl.
6.3
6.3.a
QP15 (5 points)
(1 point)
Conserving momentum gives: pi = M v0 = (2M )vf = pf v0 vf = 2 6.3.b (2 points)
The location of the new center of mass is a distance L/4 from the centers of the two rods. Using the parallel axis theorem, we have can find Irod the moment of a rod about this point.
Irod = Irod,CM + M D2 Irod =
L 1 M L2 + M ( )2 12 4 7 M L2 Irod = 48
Ic = 2Irod
Ic =
7 M L2 points)
Conserve 24
3
6.3.c
(2 angular momentum about an axis through the new center of mass at the time of the collision. Li = rcm pcm
L Li = ( )M v0 into the page 4 Li = Lf = Ic f f =
1 4 M Lv0 7 2 24 M L
f =
6 v0 7L
6.4
6.4.a
QP16 (5 points)
(2 points)
To find the direction of friction, consider the speed of the point of contact. Here, the rotation contributes at speed 2v0 to the right, in addition to the translational speed of v0 to the right. Thus the ball is sliding right, and the frictional force opposes this.
f =v /7R o vf =v /7 o
Figure 1: QP16a
M a = Ff = -M g a = -g v(t) = v0 + at v(t) = v0 - gt Considering the torques about the ball's center, and choosing > 0 to be in the direction of the ball rolling right, Newton's Law gives: I = f = RM g = RM g 2 2 5MR
5g 2R (t) = 0 + t = (t) = - The ball rolls at t0 when v(t0 ) = (t0 )R 4 5gt 2v0 + R 2R
5 v0 - gt0 = -2v0 + gt0 2 7 3v0 = gt0 2 6 v0 t0 = 7 g 6.4.b (2 points)
Using the above results, we find v(t0 ) and (t0 ). v(t0 ) = v0 - g( v(t0 ) = 6 v0 ) 7 g
1 v0 7 5g 6 v0 2v0 + ( ) (t0 ) = - R 2R 7 g (t0 ) = 1 v0 7R
>0 vo
F f
R
Figure 2: QP16b
The ball now rolls to the right with a much reduced speed. 6.4.c (1 point)
Initially, the torque of friction is into the page, with magnitude |f | = RM g as seen in part a). For t t0 , the ball rolls without slipping. The contact point is a rest, and the frictional force is zero. So while rolling, |f | = 0.
6.5
6.5.a
QP25 (5 points)
(3 points)
There are three conditions to satisfy: 1) Newton's second law, 2) the rotational analog of Newton's second law and 3) the nonslip condition. The two forces acting on the yo-yo are the tension in the string and the frictional force between the yo-yo and the floor: T + f = ma T x + f x = ma^ ^ ^ x Since it's not obvious to which direction the frictional force is acting, here assumes it's acting to the right. If it's in fact the other way round, it will be reflected by a negative sign later. If we choose the center of the yo-yo as the origin, the resulting torque equation is T + f = I
5
(-r^ T x) + (-R^ f x) = I y ^ y ^ (rT + Rf )^ = I^ z z No slipping is equivalent to (using the sign conventions above) v = -R a = -R Combining these three equations and solving for T and f yields T f 6.5.b (1 point) mR2 + I a R(R - r) mrR + I = - a R(R - r) =
(Negative sign means acting to the left)
Using the given moment of inertia I = kmR2 : |f | mg min 6.5.c (1 point) = = = mrR + kmR2 a R(R - r) mrR + kmR2 a R(R - r) r + kR a R-r g
The yo-yo rolls to the right. Motion to the left would violate the assumption of no slipping, as can be seen by checking the signs in the above equations.
6